calculus integration induction algebra polynomial function logarithms

[b]1)[/b] Show [hide="method"]by induction[/hide] that for any natural integer $n$, we have :
$e = \sum_{k=0}^{n}\frac 1{k!}+\int_{0}^{1}\frac{(1-t)^{n}e^{t}}{n!}dt$. [hide="hint"]Use an integration by parts.[/hide]
[b]2)[/b] Show that $\left|\int_{0}^{1}\frac{(1-t)^{n}e^{t}}{n!}dt \right| \leq \frac 3{(n+1)!}$. [hide="conclusion"]So $e = \sum_{n=0}^{+\infty}\frac 1{n!}$ and moreover, $\sum_{k=0}^{n}\frac 1{k!}$ is an approximation of $e$ up to $\frac 3{(n+1)!}$.[/hide]
[b]3)[/b] Give an approximate rational value of $e$ up to $10^{-3}$. [hide="note"]With the help of the previous series, one finds very quickly that $e = 2,7182818 \ldots$[/hide]

I know it's very(easy+classical), but.. anyone willing to have a try at it? :)

OK, #1 is by induction.

For $n=0,$ it readts $e=1+\int_{0}^{1}e^{t}\,dt,$ which is true.

Assume it's true for $n.$ We take the integral and integrate it by parts, integrating the polynomial and differentiating the exponential.

$\int_{0}^{1}\frac{(1-t)^{n}e^{t}}{n!}\,dt = \left.-\frac{(1-t)^{n+1}e^{t}}{(n+1)!}\right|_{0}^{1}+\int_{0}^{1}\frac{(1-t)^{n+1}e^{t}}{(n+1)!}\,dt$

The boundary terms in this integration by parts evaluate to $\frac1{(n+1)!},$ which is the next term in the sum. Hence our inductive step is proved.

[Note that what we have actually done is to prove Taylor's Theorem with the integral form of the remainder, only in this specific case.]

#2 follows from assuming that $e<3$ and replacing $e^{t}$ in the integral for the remainder by $3.$ [That begs the question: how do we already know that $e<3?$ There are several reasonable ways of answering that question.]

#3: If we want three places, it would be nice to have some control of the next place. I'll interpret this as wanting the error to be less than $5\times 10^{-4}.$ From #3, we need

$\frac3{(n+1)!}<5\times 10^{-4},$ or $(n+1)!>6000.$ This is satisfied for $n=7.$ From there,

$\sum_{k=0}^{7}\frac1{k!}=\frac{13700}{5040}\approx 2.718.$

#4? That was my question; I'll let someone else answer it. Hint: if $e=\frac{p}{q}$ in lowest form, what happens if you let $n=q$ in #1 and multiply by $q!?$

Thank you! :)

Hmm, for #3, what I had done was :
$|e-\sum_{k=0}^{n}\frac 1{k!}| \leq 10^{-3}\Longleftarrow \frac 3{(n+1)!}\leq 10^{-3}\iff (n+1)! \leq 3000 \Longleftarrow n = 6$
So I had :
$\sum_{k=0}^{6}\frac 1{k!}= \frac{1957}{720}= 2,718\ldots$.

Is it also a correct interpretation? (I know it's a dumber interpretation.. but I just want to know if it's also right.)

Yes, that is probably also a correct interpretation.

Ok, thank you.

[quote="Kent Merryfield"]...

[That begs the question: how do we already know that $e<3?$ There are several reasonable ways of answering that question.]

...[/quote]
As you said, I think there are many different ways to answer that question.
I guess it depends on how the function $\exp$ was introduced.
Namely, if one sees it as the function satisfying $f' = f$ (and $f(0) = 1$), I don't think it helps much.
However, if we first see its reciprocal function, $\ln$, since it's defined as $\int_{1}^{t}\frac{dt}t$, then, using, for example, the method of trapezoids, one can get some bounds on $e$.

I don't know if I was clear enough though..

[quote="Kent Merryfield"][b]4)[/b] Use this to prove that $e$ is irrational.[/quote]
It looks like nobody wants to do this, so..
Let's assume that there exist $(p, q) \in \mathbb{N}^{2}$ such that $e = \frac{p}{q}.$
$\forall n \in \mathbb{N}, \ \ 0 < \frac{p}{q}-\sum_{k=0}^{n}\frac 1{k!}< \frac 3{(n+1)!}$
or,
$\forall n \in \mathbb{N}, \ \ 0 < \underbrace{pn!-q\sum_{k=0}^{n}\frac{n!}{k!}}_{\in \mathbb{N}}\leq \frac{3q}{n+1}$
Especially, for $n = 3q, \ \ 0 < \text{integer}\leq \frac{3q}{3q+1}< 1,$ so contradiction and conclusion.


Let $ P_1$, $ P_2$, $ P_3$, $ P_4$ be five distinct points on a circle. The distance of $ P$ from the line $ P_iP_k$ is denoted by $ d_{ik}$. Prove that $ d_{12}d_{34} \equal{} d_{13}d_{24}$.

Denote by $ M_1,M_2,M_3,M_4$ the feet of the altitudes from $ P$ to $ P_1P_2,P_2P_3,P_3P_4,P_4P_1$. The conclusion results from the fact that $ \triangle PM_1M_4 \sim \triangle PM_2M_3$ which can be easily deduced, only using some cyclic quadrilaterals and some angles.

Complex numbers:
If $a,b,c$ lie on a unit circle centered in $0$ then $dist(A,BC)=\frac{|a-c||a-b|}{2}$
Proof: let $x$ be a foot of $A$ on $BC$. Then $2x=b+c+a-\frac{bc}{a}\implies |x-a|=|\frac{b+c-a}{2}-\frac{bc}{2a}|=|\frac{ab+ac-a^2-bc}{2a}|=\frac{|b-a||a-c|}{2}$
Now take points $P,P_1,...,P_4$ on Argand plane such that they lie on a unit circle centered in $0$.
$d_{12}d_{34}=\frac{|p-p_1||p-p_2||p-p_3||p-p_4|}{4}$
$d_{13}d_{24}=\frac{|p-p_1||p-p_3||p-p_2||p-p_4|}{4}$


Is there  a continuous function so that $ f(\frac {m_1}{n_1})\equal{}irrational$ and $ f(irrational)\equal{} \frac {m_2}{n_2}$
where $ m_1,m_2,n_1,n_2 \in Q$  ?

[quote="amparvardi"]Is there  a continuous function so that $ f(\frac {m_1}{n_1}) \equal{} irrational$ and $ f(irrational) \equal{} \frac {m_2}{n_2}$
where $ m_1,m_2,n_1,n_2 \in Q$  ?[/quote]

No : $ f(x)\minus{}x$ would be a continuous function whose all values are irrational, and so constant. So $ f(x)\equal{}x\plus{}a$
But then $ f(0)\equal{}a$ and so $ a\notin\mathbb Q$ and then $ f(a)\equal{}2a$ should be rational, which is wrong.


$0.5(cos5x+cos7x)-cos^2(2x)+sin^2(3x)=0$
Solve for x?

Using the trigonometric formulas

$(1) \; 2\cos^{2}u = 1+\cos2u$

$(2) \; 2\sin^{2}u = 1-\cos2u$

$(3) \; 2\cos u \cos v = \cos(u-v)+\cos(u+v)$

$(4) \; 2\sin u \sin v = \cos(u-v)-\cos(u+v)$

we obtain the following series of equivalences:

$0.5(\cos5x+\cos7x)-\cos^{2}2x+\sin^{2}3x = 0$

$(\cos5x+\cos7x)-2\cos^{2}2x+2\sin^{2}3x = 0$

$2\cos x \cos6x-(\cos4x-1)+(1-\cos6x) = 0\;\;\;$ (from (3), (1) and (2))

$2\cos x \cos6x-(\cos4x+\cos6x) = 0$

$2\cos x \cos6x-2\cos x \cos5x = 0\;\;\;$ (from (3))

$2\cos x(\cos6x-\cos5x) = 0$

$\cos x \sin({\textstyle \frac{11}{2}x})\sin({\textstyle \frac{1}{2}x}) = 0$

$x ={\textstyle \frac{\pi}{2}+k\pi, \; \frac{2k}{11}\pi, \: 2k\pi \;\;\; (k}$ arbitrary integer)


geometry
combinatorics proposed
combinatorics

baron has a garden.of apples and oranges and we know that around each apple tree  with the radius $10$ there exists exactly $10$ orange tree.
baron say that the number of the orange trees is less! than the apple trees.is this possible?

hmmmm.i 've just proved if baron say the truth then the number of  apple trees is greater than$45$.

isn't there any one wo has solved it? :?:

I think there's something missing in the problem statement.
Nothing stops me from putting a thousand apple trees and ten orange trees in a very small area which makes the statement trivially true.

So maybe two trees have to have at least distance 1?

In that case, what's wrong with the arrangement:
        a
       a a
     a o a
    a o o a
  a o o o a
 a o o o o a
a a a a a a a

Or maybe in the radius 10 area, there are not allowed to be any other apple trees?

i think you'v misunderstood sth in problem :D .

In case you did not notice: I was asking for a clarification.


calculus
integration
real analysis
real analysis unsolved

how do we obtain the formula e for[size=150][b] integral (sin)^n dx from 0 to pie/2 . [/b][/size]

please put the solution

http://www.math.niu.edu/~richard/Math230/reduction.pdf

It's a standard exercise, hope this helps.


Prove that the following equation has only one solution
$\sqrt[n]{3x+2}+\sqrt[n]{5x+4}=\sqrt[n]{2x+3}+\sqrt[n]{4x+5}$. Any idea is wellcome(if it uses integrals I might need a little explantion)! Thank you :)

huh? Corect me if I`m wrong but if we take $n\in [0;1)$ all x works(with the condition that the numbers under the square root must be positive.

NO. sorry but [n] is the order of the sqrt. I don't know why $LaTeX$ put it that way... Edited now.

[quote="andyciup"]Aah..I get it.. Sorry for the misunderstanding..we are talking integers right?(both n and x)[/quote]

Only $n\geq2$ is integer.

Anyone? Could anyone show that if the 2 sides are 2 funtions(increasing) they met in one point..? A hint would be usefull:)

For positive $x$ it is not hard: show that if $x$ is a solution, so is $\frac1x$ and show that there are no solutions larger than 1.

In fact, I think we can generalize to a broader range of $n$, if we wanted.


By A-G Inequality, we have :

$ 1\plus{}\frac{3}{a\plus{}b\plus{}c} \ge \frac{2\sqrt{3}}{\sqrt{a\plus{}b\plus{}c}}\equal{}\ge \frac{2\sqrt{3}}{\sqrt{abc(a\plus{}b\plus{}c)}}$

So we just need to prove that :

$ ab\plus{}bc\plus{}ca \ge \sqrt{3abc(a\plus{}b\plus{}c)}$

or

$ \sum_{cyc} a^2(b\minus{}c)^2 \ge 0$, which is true  :wink:

[quote="amparvardi"]Hi everybody.
solve this problem please:
We know That $ abc \equal{} 1$.prove that:
$ 1 \plus{} \frac {3}{a \plus{} b \plus{} c}\geq \frac {6}{ab \plus{} bc \plus{} ca}$[/quote]
It is  [b]Romania Junior TST 2003[/b]. :)


Find all functions $ f: (0,\infty) \to (0,\infty)$ , $ f(f(x)) \plus{} f(x) \equal{} 6x$

[quote="pardesi"]ur topic says it's 6x while question 2x.which is correct??[/quote]

I m sorry the correct form is with $ 6x$ . I edited.

[hide]
For an arbitrary positive real number $ a$, define $ a_k$ by $ a_0 \equal{} a,a_{k \plus{} 1} \equal{} f(a_k)$. Then $ a_{k \plus{} 2} \plus{} a_{k \plus{} 1} \equal{} 6a_k$. Verify that
\[ a_{k \plus{} 2} \minus{} 2a_{k \plus{} 1} \equal{} ( \minus{} 3)(a_{k \plus{} 1} \minus{} 2a_k)\Rightarrow a_{k \plus{} 1} \minus{} 2a_k \equal{} ( \minus{} 3)^k(a_1 \minus{} 2a_0)
\]

\[ a_{k \plus{} 2} \plus{} 3a_{k \plus{} 1} \equal{} 2(a_{k \plus{} 1} \plus{} 3a_k)\Rightarrow a_{k \plus{} 1} \plus{} 3a_k \equal{} 2^k(a_1 \plus{} 3a_0)
\]

\[ a_k \equal{} \frac {2^k(a_1 \plus{} 3a_0) \minus{} ( \minus{} 3)^k(a_1 \minus{} 2a_0)}{5}
\]
But $ a_k$ is positive for all $ k$, so this can only happen if $ a_1 \minus{} 2a_0 \equal{} 0$. (If it were positive, there would be a problem for large enough even $ k$, and if it were negative, there would be a problem for large enough odd $ k$.) This means $ f(a) \equal{} 2a$. Since this is true for all positive $ a$, the only function is $ \boxed{f(x) \equal{} 2x}$.
[/hide]


MATHCOUNTS
blogs
geometry
AMC
USA(J)MO
USAMO
Pythagorean Theorem

I have no chance, unless my practice amazing hard...and get a bit lucky, i think at least one person's gonna make to nats from my school though..our state, VA, has it later.

Only one person---me! BTW, no one else in my school's history has ever even made it to chapter countdown. :(

Is your chapter exceedingly hard or is your school not motivated to do competitions?

My chapter hard? Not at all! My school only had 3 years in its history of MATHCOUNTS participation where we had full 4-person teams. :(

My school had 3 years, 2 being one person and only one having 4 people. I was the first person make chapter count, states, state cd, and nats...

My school has had a couple people make nats CD in the past but none will make nats this year.

honestly, how many schools could get 4 students into nats?
that would take a REALLY GOOD school (top 4 all 42+) or a good school in a crappy state

I think Missouri did that this year? (I'm not sure.....) :blush:
By the way, who's that person that clicked "4 person"  on the poll?

missouri.. duh... mewto, math, tinytim, and someone else... i dunno.

Missouri did it.

mewto, tinytim, math154, dragonlaird7.

Yeah, I clicked it in the poll.

xpmath was correct about the MO team:

Mewto55555
tinytim
math154
dragonlaird7 (me)

And we're not just from the same school- we're all from the same lunch table.

Whoa, how did you post twice in the same minute with different messages?

Anyway, I'm laughing how Mewto sits with his sister at lunch :rotfl:

sorry mew and dragonlaird :)

Or 2 :D

Which is not entirely impossible, considering that the team being discussed has another year.

[quote="nikeballa96"]one school, a legendary school in MI (echen, kyyuanmathcount (who got second at nats.), and many others go there) screwed up their individual schools at chapters, so didn't qualify for states, [/quote]

Well my school relies on a hard team round which we always do well on and other teams fail which usually boosts us. Unfortunately the team round was amazingly simple and we all failed written. Also I did not make chapter CD (stupid mistakes killed me) but made state CD (stupid mistakes still killed me).

AIME15: on the "that team has 3 USAMOers" comment, CA had 4 USAMOers the year before but failed epicly.

I got something like 5th school, 4th chapter, 3rd state, so 2nd national? No, make that 1st.

Bobby will pwn you in cd.

5th school? Lulz.  That's amusing.

blargh, that's why I hate countdown

but written I can do fine

Fine, 2nd is OK with me

[quote="james4l"]And the might-be moppers are the seventh graders...

It would be funny if a state team had an IMOer.[/quote]
cant seventh graders only make black mop? if eight graders can go to red mop, i will be very happy  :D

Hey james. I'm sure you're like beastly, but don't count your eggs before they hatch. Cause you might fail like I did cause I was sooo nervous. And get 7th. And the dream dies. Which really stinks because then you feel bad and you're embarrassed because everyone already expected you would win. Like me.  :(

Anyone can make blue MOP, so seventh graders can make blue and don't have to make black.

Look here, in reality I am happy to get 8th at Nats (quarterfinalist).
But that's fail.

Zz...Zz: blue MOP is reserved for non-seniors, so "anyone" can make blue MOP is not entirely true.

I'm pretty sure pythag011 got into at least blue MOP.

Red MOP is reserved for freshmen = not 8th graders.

[quote="james4l"]*snip*
Red MOP is reserved for freshmen = not 8th graders.[/quote]
darn. that's not fair. oh well.

but you get the opportunity next year.

That's somewhat their rationale for reserving it for 9th graders.

yesh. the goal: make the cutoff this year, secure a spot next year

Wow, look how far we drifted off topic!

Anyways, USAMO definitely fluctuates in difficulty each year. So, Red MOP this year does not mean Red MOP next year.

It's a surprise I didn't make nats though. I'm missing out so much experience.

Those two sentences have nothing to do with each other.


function
trigonometry
algebra unsolved
algebra

Let f(x): $ f(1\minus{}x)\plus{}2f(x)\equal{}sin^2x$
Prove that :$ sinf(x)<\frac{1}{\sqrt{2}}$

[quote="karis"]Let f(x): $ f(1 \minus{} x) \plus{} 2f(x) \equal{} sin^2x$
Prove that :$ sinf(x) < \frac {1}{\sqrt {2}}$[/quote]

$ P(x)$ : $ f(1 \minus{} x) \plus{} 2f(x) \equal{} \sin^2x$
$ P(1\minus{}x)$ : $ f(x) \plus{} 2f(1\minus{}x) \equal{} \sin^2(1\minus{}x)$

Eliminating $ f(1\minus{}x)$ between these two equations gives : $ f(x)\equal{}\frac{2}{3}\sin^2x\minus{}\frac{1}{3}\sin^2(1\minus{}x)$

So $ f(x)\in[\minus{}\frac{1}{3},\frac{2}{3}] \subseteq [\minus{}\pi,\frac{\pi}{4})$

Hence the result.


superior algebra
superior algebra solved

If $\alpha \in C,$ $\alpha^{n} \neq 1,$ $A \in M_{n}(C)$ and $A^{T} = \alpha A$ compute $det(A).$

cheers! :D

$det(A^T)=det(\alpha A)$
$det(A)=\alpha^ndet(A)$
$(\alpha^n - 1)det(A)=0
$det(A)=0$


AMC
USA(J)MO
USAMO
induction
number theory
prime numbers
relatively prime

My solution to #1 according to my rough draft, approximate scores are appreciated!:

Answer: All composite integers satisty the conditions except when $n=pq$ where p, q are prime integers.

Proof: When $n=pq$ for prime numbers $p, q, n$ has 3 divisors: $p, q, pq$. Since $p, q$ will always be adjacent and $p, q$ are relatively prime, contradiction.

We will show that every other composite integers will sastify the condition through the following 2 cases

Case I : If $n= p^a$, where $p$ is a prime integer and $a \geq 1$. Note that every divisor of n will be divisible by p, so none of the divisors can be relatively prime.

Case II: All integers $n=p_1^{a_1} *p_2^{a_2}*...p_k^{a_k}$, excluding the two cases above, can be arranged on such circle.
   Proof: We will prove this by induction.
When $k=2$, then we can arrange like in this order:
$p_1, p_1^2,... p_1^{a_1},  p_1p_2,  p_2,  p_2^2,...  p_2^{a_2}$, rest of divisors arranged in any order(since all of the rest of the divisors of n are divisible by $p_1p_2$).

Now assume if it's possible to arrange the divisors of $n=p_1^{a_1} *p_2^{a_2}*...p_k^{a_k}$, consider $np_{k+1}^{a_{k+1}}$. We can rearrange as below(in that order):

$d_1p_{k+1}, d_1, d_2, ... d_x$(where d_i are all divisors of n, we assumed such arrangement is possible), $d_xp_{k+1}$, rest of divisors of $np_{k+1}^{a_{k+1}}$ in any order(all of these are divisible by $p_{k+1}$).
(on USAMO I had a diagram with left side $d_i$'s, right side "rest of the divisors", I didn't explicitly say how to connect left and right although it's trivial)
Induction is complete
QED

[quote="qvc122"]Don’t make yourself craze. You will get your score soon. Looks like you did a reasonable job. As a freshman, you have 98% of chance to be included in the MOP.[/quote]

Ah looks like this year a low nonzero score may not be enough for Red MOP... :( and I know that at the very end of #4 I made a careless mistake that I got the wrong answer  :((so there's a good chance I'll get a 2/7 on #4 even though I had eveything else right), so this problem is my only hope...

Stop worrying beta.  I am 99.999% sure that a 7+ score will get you in.  Although you didn't need to induct, I think this solution should give you almost certainly a 7 and certainly 6+ if the grader is somehow mean.  Also, you should get at least 3 on your other solution (by what you've said, anyway).

Wait! How did you figure out how many points you got on each question?

[quote="fuzzylogic"][quote="beta"]Apparently this proof deserved only 2 pts, can someone tell what the major error in this proof is?[/quote]

How do you arrange $pqr$? Based on your induction argument, you assumed that $pq$ can be arranged.[/quote]

Thanks, ahhh I'm so stupid!

I'm not sure I understand where the d's came from, and I believe for case 3 you only showed that it worked when in a straight line.  But of course you did not copy your proof word for word here.


The following problem is displayed  [b][u]word for word[/u][/b]
a certain pomegranite has red and blue seed and each color accounts for exactly one half of the twenty seeds in the pomegranite if kee young picks four seeds from the pomegranite and eats each seed immeadiatly after picking them what is the probability that two seeds of each color end up in his stomach?
i said the ans is $ \frac{10*10*9*9*(4C2 or 6)}{20*19*18*17}$=$ \frac{135}{323}$ or is it $ \frac{10*10*9*9}{20*19*18*17}$=$ \frac{45}{646}$ as the school would say :?:

$ 135/323$ is correct. Of the five possibilities 4 red, 3 red, 2 red, 1 red, 0 red, 2 is the most likely, so that has to have at least a $ 1/5$ chance. The second answer can't be right.

Here's how I thought about it, may be wrong considering how big your answers are...

He can pick the first two in any way. He could either pick them both of the same color, or of different colors.

Same color: He needs to pick the remaining seeds of the same color. He can do this with probability $ \frac{10 * 9}{18 * 17} \equal{} \frac{5}{17}$.

Different colors: He can pick the third in any way. The remaining seed must be of the color he didn't yet choose, with prob $ \frac{9}{17}$.

As both cases have equal probability, the total probability is the average between the two: $ \frac{\frac{9}{17} \plus{} \frac{5}{17}}{2} \equal{} \frac{7}{17}$.

But if he eats it there is a 99.995% chance it will end up in his stomach, so you have to multiply by that  :D

exactly, the half probability varies as the seeds are picked. i agree with the first answer.

[quote="MellowMelon"]ffao, there are not equal chances of choosing the same color or different colors, so that doesn't work. You can fix it by computing the probabilities of getting same and different colors and doing a weighted average.[/quote]

Aha, I knew I did something wrong. The chance of getting same colors is, in fact, $ \frac {9}{19}$ (not $ \frac {1}{2}$), so the correct answer is,

$ \frac {9 * 5 \plus{} 10 * 9}{19 * 17} \equal{} \frac {135}{323}$

[quote="peregrinefalcon88"]The following problem is displayed  [b][u]word for word[/u][/b]
a certain pomegranite has red and blue seed and each color accounts for exactly one half of the twenty seeds in the pomegranite if kee young picks four seeds from the pomegranite and eats each seed immeadiatly after picking them what is the probability that two seeds of each color end up in his stomach?
i said the ans is $ \frac {10*10*9*9*(4C2 or 6)}{20*19*18*17}$=$ \frac {135}{323}$ or is it $ \frac {10*10*9*9}{20*19*18*17}$=$ \frac {45}{646}$ as the school would say :?:[/quote]

Shouldn't it be 10C2 for the red and 10C2 for the blue seeds?  So the school's solution should be right.

[quote="cdymdcool"]
Shouldn't it be 10C2 for the red and 10C2 for the blue seeds?  So the school's solution should be right.[/quote]

If you're counting like that, the outcome space would be all possible ways to choose 4 seeds from 20 and the denominator would be $ \binom{20}{4}$, not $ 20\times 19\times 18\times 17$. The probability works out to peregrinefalcon's answer.

I agree with peregrine falcon's answer.

I checked both ways and the computation.

[hide="a solution"]The total number of ways to pick 4 seeds from 20 is $ \binom{20}{4}$.

The total number of ways to pick 2 red seeds from 10 is $ \binom{10}{2}$, and this also applies for blue seeds. Thus the total number of ways to pick 2 blue seeds and 2 red seeds is $ \binom{10}{2}^2$.

The probability that he picks 2 blue and 2 red is $ \frac{\binom{10}{2}^2}{\binom{20}{4}} \equal{} \frac{135}{323}$.

 [/hide]


hey, i got both book + both answer books
its 70 something value. Mine are in perfect condition.. would sell this on ebay but i doubt anyone on there would buy it.. i barley used them i got lazy. no writing no damage at all. please email me at kentliu@gmail.com for offers.. thanks

[quote="bubala"]Can you legally do this w/o AoPS's permission?  Idk...[/quote]
 :? Can you explain it??

[quote="Danbert"]Why wouldn't you be able to sell your books without the permission of the publisher?  After all, you do own them...[/quote]
if you look on the back of juice boxes, it will say something like "not for individual resale" or something like that

if you'll buy my shipping cost to istanbul, maybe i'll buy them

70 dollars... and then send to Beijing...
i don't think it will be good for me if i really buy them.

[quote="MithsApprentice"][quote="Scrambled"]if you look on the back of juice boxes, it will say something like "not for individual resale" or something like that[/quote]

That has nothing to do with used books. You can definitely sell your AoPS books if someone is willing to buy them, not that I recommend doing so...[/quote]

Yeah.  Besides, the juice box thing is for when they're in a package of, say, 12.  Then you can't sell them one by one.

[quote="filletwho"]anyone one have art and craft they want to sell?[/quote]OK let's not make this a marketplace thread. Anyone wanting to buy kentliu's books, just PM or e-mail him. Thread closed.


Consider the circle whose center is in the 1st quadrant and which is tangent to both coordinate axes and the line L whose equation is: $3x+4y=120$.  Determine the coordinates of the point of tangency of the circle and line L.

Let the line 3x + 4y = 120 cut the x-axis at a and y-axis at B.
a = OA = 40, b = OB = 30, c = AB = 50,
s = semiperimeter of triangle OAB = 60
radius of the circle = r = s - c = 10
center C = (r, r) = (10, 10)
Equation of the circle is
(x - r)^2 + (y - r)^2 = r^2
x^2 + y^2 - 2r(x + y) + r^2 = 0
x^2 + y^2 - 20x - 20y +100 = 0
Line L' through C(10, 10) perpendicular to L is 4x - 3y = k = 40 - 30 = 10
Point of tangency of L and the circle is the point of intersection of L and L'
Solving 3x + 4y = 120 and 4x - 3y = 10 we get x = 16, y = 18
Pt of tangency is (16, 18)


trigonometry
induction
Trigonometric Identities
algebra
IMO
IMO 1966

Prove that for every natural number $n$, and for every real number $x \neq \frac{k\pi}{2^t}$ ($t=0,1, \dots, n$; $k$ any integer) \[ \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}  \]

[b]Solution.[/b] By the double angle formula for cosine, $ \cos 2^{k}x\equal{}2\cos^{2}2^{k\minus{}1}x\minus{}1$.  Dividing both sides by $ \sin 2^{k}x$ ($ \neq 0$ by the given values of $ x$),
\[ \cot 2^{k}x\equal{}\cot 2^{k\minus{}1}x\minus{}\frac{1}{\sin 2^{k}x}\implies\frac{1}{\sin 2^{k}x}\equal{}\cot 2^{k\minus{}1}x\minus{}\cot 2^{k}x.\]
Hence,
\[ \sum_{k\equal{}1}^{n}\frac{1}{\sin 2^{k}x}\equal{}\sum_{k\equal{}1}^{n}{(\cot 2^{k\minus{}1}x\minus{}\cot 2^{k}x)}\equal{}\cot x\minus{}\cot 2^{n}x.\]

And the conclusion follows. $ \Box$

I understand that the difficulties of old IMO problems are lower, but this is a bit ridiculous. Even if you don't see the telescoping solution, you can just use induction.

It is a very nice problem though.

Sorry to revive, but here's another solution.

[hide="Induction Solution"]
Base case, $n=1$ gives $\frac{1}{\sin{2x}}=\cot{x}-\cot{2{x}}$, which is true: $\cot{x}-\cot{2x}=\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}=\frac{\cos^2{x}}{\sin{2x}}-\frac{\cos^2{x}-1}{\sin{2x}}=\frac{1}{\sin{2x}}$.

Assume that the identity holds for $n=k$. For $n=k+1$, we get

$\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\cdots+\frac{1}{\sin{2^k x}}+\frac{1}{\sin{2^{k+1}x}}$

$=\cot{x}-\cot{2^k{x}}+\frac{1}{\sin{2^{k+1}x}}$

$=\cot{x}-\frac{\cos{2^k x}}{\sin{2^k x}}+\frac{1}{2\sin{2^kx}\cos{2^kx}}$

$=\cot{x}-\frac{2(\cos{2^kx})^2}{2\sin{2^kx}\cos{2^kx}}+\frac{1}{2\sin{2^kx}\cos{2^kx}}$

$=\cot{x}-\frac{2(cos{2^kx})^2-1}{2\sin{2^kx}\cos{2^kx}}$

$=\cot{x}-\frac{\cos{2^{k+1}x}}{\sin{2^{k+1}x}}$

$=\cot{x}-\cot{2^{k+1}}$

So the identity holds for $n=k+1$, completing the induction.[/hide]

Also used induction.

[hide=sol]
We can use induction. The base case is $n=1.$ We get $\frac{1}{\sin(2x)} = \cot x - \cot(2x)$. Let's simplify the RHS first. To make it easier, we can write this in terms of $\tan$ and use the double angle formula to get $\frac{1}{\tan x} - \frac{1-\tan^2 x}{2\tan x} = \frac{\tan^2 x +1}{2\tan x}.$ Writing this in terms of $\sin$ and $\cos$, we get $$\frac{\frac{\sin^2 x}{\cos^2 x}+1}{\frac{2\sin x}{\cos x}} = \frac{\sin^2 x + \cos^2 x}{2\sin x \cos x } = \frac{1}{2 \sin x \cos x}.$$ But $\frac{1}{\sin(2x)} = \frac{1}{2\sin x \cos x}$. We are done. Now for the inductive step. We need to show that $\frac{1}{\sin{2^{n+1}x}} = \cot(2^n x) - \cot(2^{n+1}x).$ Again, we first work with the RHS. We can write this as (using double angle formula) $$\frac{1}{\tan(2^n x)} - \frac{1-\tan^2 (2^n x)}{2\tan(2^n x)} = \frac{\tan^2 (2^n x)+1}{2\tan(2^n x)}.$$ Again, we can write this in terms of $\sin$ and $\cos$. We get $$\frac{\frac{\sin^2 (2^n x)}{\cos^2 (2^n x)}+1}{\frac{2\sin(2^n x)}{\cos(2^n x)}} = \frac{\sin^2 (2^n x) + \cos^2(2^n x)}{2\sin (2^n x) \cos(2^n x)} = \frac{1}{2\sin (2^n x) \cos(2^n x)}.$$ But $\frac{1}{\sin(2^{n+1} x)} = \frac{1}{2\sin (2^n x) \cos (2^n x)}.$ We are done with the inductive step, so we are done.
[/hide]


How is the alcumus score determined? Is there a formula?

Yes there is a formula. It is very complex with its many factors. The formula will not be revealed to users, but considers difficulty of problems solved, percentage correct, etc. Just try your best and let AoPS handle it.

Oh, thanks. :)

But what is the max Alcumus Score you can get? Or is there no limits?

The max alcumus score is most likely 100, but it is nearly impossible to get.

The only way to get that would be to get every single question correct on your first try every single time, and even then, it might not be 100 due to other factors.

[quote="Fidar"]Yeah, I'm pretty sure that you'd have to answer every one correct on your first try, because it says on the report page: "the alcumus score is an approximation of the number of problems in the alcumus database that you are expected to get correct."
(Number of problems=percentage)[/quote]
Number of problems isn't your percentage. It's based on how well you do on the problems in the Alcumus database compared to others rather than just a percentage of how much you did. You're percentage is shown in another corner-the right upper corner. Your Alcumus score has no effect on the percentage. If some people did bad on a few problems you did right, then you get more score.

[quote="Fidar"]what I meant by percentage is not the problems you have already done
I meant the the total number of problems in the database of alcumus[/quote]

To make things more clear about Alcumus score:
The Alcumus database itself has WAY more than 100 problems. It has over 1100 problems for you to do, and the percentage of how many you have gotten right is at the right hand corner. For the Alcumus score, it is more of a comparason score between you and other people who use Alcumus. If you use it often and do problems correctly that others may do incorrectly, your Alcumus score will go up more. If you do many problems, but get most incorrect, your Alcumus score will go down. If you do many problems, and get about 1/2 right, your Alcumus score will most likely go up a little. If you get many hard problems correct, your Alcumus score will shoot higher. If you get a problem wrong in a section which you have passed/mastered/done already, your Alcumus score will fall heavely. Alcumus score is not updated in real time. Your percentage is just a percentage of how many problems you get right out of how many you do totally. Therefore, your Alcumus score is more dependent on the factor of the problems you get right, the difficulty, and comparason between your report and others.

I solved 36 problems and had one or two that I got on the second try and I finally got 100% :)


An optically active compound X has the molecular formula $ \ce{C6H10}$, and it contains no triple bond. When catalytic hydrogenation is carryied out it yields a compound with molecular formula $ \ce{C6H14}$, which is optically inactive. Propose a reasonable structure for X.

hex-2,3-diene ?

this would be an allene: two neighboring double bonds, i.e. two orthogonal $ \pi$ systems that cannot rotate past each other and hence gives rise to chirality. On one end of the allene are a methyl and a hydrogen and on the other side are an ethyl and a hydrogen.


algorithm
algebra
polynomial
number theory
Euclidean algorithm
greatest common divisor
relatively prime

Show that for any positive integer $ a$, $ \frac{a^3\plus{}2a}{a^4\plus{}3a^2\plus{}1}$ is always a fraction in lowest terms.

The two ways I know for finding common divisors are Euclidean Algorithm and factoring.  I don't see how to factor the bottom, and using the Euclidean Algorithm doesn't work because there's no like terms.  Help!

Denote by $ (a,b)$ the greatest common divisor of $ a$ and $ b$.

[hide="Solution"]
We must show that $ (a^3\plus{}2a,a^4\plus{}3a^2\plus{}1)\equal{}1$.
By subtracting $ a^3\plus{}2a$ from $ a^4\plus{}3a^2\plus{}1$, $ a$ times, we have $ (a^3\plus{}2a,a^2\plus{}1)$.
By subtracting $ a^2\plus{}1$ from $ a^3\plus{}2a$, $ a$ times, we have $ (a,a^2\plus{}1)$.
Now, subtract $ a$ from $ a^2\plus{}1$, $ a$ times, and we have $ (a,1)$, which is clearly 1, and we are done.[/hide]

It is a well known rule that $ (a,b)\equal{}(a\minus{}b,b)\equal{}(a,b\minus{}a)\equal{}(a\plus{}b,b)$, etc.

The thing about using the Euclidean algorithm with polynomials in $ x$ is that you are allowed to multiply by $ x$ - this is an essential step in the division algorithm, so it's an essential step in properly using the Euclidean algorithm as well.

The other thing about polynomials is that showing that two polynomials have no common [i]polynomial[/i] factors is not enough to show that their integer values will always be relatively prime; this is clearly not true of, for example, $ x$ and $ x \plus{} 2$.  In other words,

[i]Given polynomials $ f(x), g(x)$ with integer coefficients, the condition that there exist polynomials $ p(x), q(x)$ with integer coefficients such that

$ f(x) p(x) \plus{} g(x) q(x) \equal{} 1$

(which is what the Euclidean algorithm lets you do) is strictly stronger than the condition that $ \gcd( f(x), g(x)) \equal{} 1$.  The two conditions are equivalent if integer is replaced by [b]rational[/b].[/i]


I am sorry. I posted this at the wrong place. I thought I posted on Others->Fun And Factory Games but somehow it ended it. Probably because I switch screen accidentally. Moderator: Could you move the post? Thanks.


1) In square ABCD with side 10, arcs AXC and AYC are formed by circles of radius 10 centered at D and B, respectively. Find the shaded Area.

2) Circle K and Circle C, both with Radius 12, intersect in H and W. If HW is 12, find the area of intersection of Circles K and C.

[hide="Hints for both questions"]
Use right triangles and the area of each circle to find the area of the intersection.[/hide]
Let me know if this doesn't help.

[hide="1"]

The area of sector ACD is one quarter the area of the circle, so it's $25\pi$. Thus, the area of region ABC is $100-25\pi$. Similarly, the area of region ADC is $100-25\pi$. Subtract this from the area of the whole square to get the area of the football, $100-(200-50\pi)=50\pi-100$.[/hide]

thanks. the hint realllllly helped. and i checked from the solution.

wow nice hint

I found that you can draw a line through the intersection and subtract from the sector of the circle. and then add both  arcs together.

number one had 90 degree sectors, number 2 had 60 degree sectors


linear algebra
matrix
superior algebra
superior algebra unsolved

Hm if F=F5 what is the order of the conjugacy class of {{1,0},{0,2}} matrix ((1,0) and (0,2) are two rows) in GL2(F5).
Do you think I should first do the order of GL2(F5) and then go from there?

the centralizer of the given matrix is $ C: \equal{}C_{GL(2,5)}\left({1\;\;0\atop0\;\;2}\right)\equal{}\left\{\,\left(\left.{a\;\;0\atop0\;\;b}\right)\,\right|\,a,b\in\mathbb{F}_5^*\right\}\simeq\mathbb{F}_5^*\times\mathbb{F}_5^*$, i.e. the order of the conjugacy class is $ \frac{|GL(2,5)|}{|C|}\equal{}\frac{(5^2\minus{}1)(5^2\minus{}5)}{16}\equal{}30$


Prove that if p and q are adjacent odd primes (ie. 3,5), then their sum p+q necessarily has 3 prime factors.  The factors need not be distinct.

WLOG p<q

First, since p and q are both odd, their sum must be even. We can thus conclude that one of the factors of (p+q) is two. In addition, (p+q)/2 cannot be prime, because p<(p+q)/2<q, and if (p+q)/2 was prime, p and q would not be adjacent primes. Thus, p+q must have at least 3 non-distinct prime factors.

the original problem stated that the primes both had to be odd

[quote="drunner2007"][quote="computer_nuke"][quote="drunner2007"]Let $p$ and $q$ be adjacent primes. We can write $p=6k+1$ and $q=6k-1$ (except for the case where $p=5, q=3$, which is easilly verified). So $p+q=12k$, which satisifies what you're trying to show.

By the way, it doesn't work with $p=3$, $q=2$, so it would be better to say [i]twin[/i] primes.[/quote]

Why 6k+1? All primes except 2 can be represented in the form 4n+1 or 4n-1.[/quote]

Yeah because if an integer is of the form $4n+1$ or $4n-1$, then it is congruent to 1 or 3 modulo 4, which is basically saying that all primes are odd.

6k+1 is a lot stronger. the more specific the form, the stronger the result (in this case using multiples of a larger number makes the form more "specific")[/quote]

Sorry if I'm being stupid and not seeing something, but since 2 is a prime, wouldn't 6k+1=2 yield 1/6 for k? and if 6k+1=3, k=1/3. Sorry, please explain.

[quote="computer_nuke"][/quote][quote="drunner2007"][quote="computer_nuke"][quote="drunner2007"]Let $p$ and $q$ be adjacent primes. We can write $p=6k+1$ and $q=6k-1$ (except for the case where $p=5, q=3$, which is easilly verified). So $p+q=12k$, which satisifies what you're trying to show.

By the way, it doesn't work with $p=3$, $q=2$, so it would be better to say [i]twin[/i] primes.[/quote]

Why 6k+1? All primes except 2 can be represented in the form 4n+1 or 4n-1.[/quote]

Yeah because if an integer is of the form $4n+1$ or $4n-1$, then it is congruent to 1 or 3 modulo 4, which is basically saying that all primes are odd.

6k+1 is a lot stronger. the more specific the form, the stronger the result (in this case using multiples of a larger number makes the form more "specific")[/quote][quote="computer_nuke"]

Sorry if I'm being stupid and not seeing something, but since 2 is a prime, wouldn't 6k+1=2 yield 1/6 for k? and if 6k+1=3, k=1/3. Sorry, please explain.[/quote]

the general form $6k-1$ or $6k+1$ applies for primes greater than 3. in my solution, i made sure to mention the other cases explicitly

[quote="mna851"]You can't say that $p=6k-1$ and $\ q=6k+1$ for this problem, because that only holds for adjacent primes that differ by $2$.  What about sets of primes such as $31$ and $37$?  However, you can say that $p=6k_{1}\pm 1$ and that $q=6k_{2}\pm 1$.[/quote]

ok good point. does the original problem ask for adjacent or twin primes?

Adjacent, which I assumed to mean $p_{i}$ and $p_{i+1}$.  123s solution is I think the standard solution.

No it's still right, the problem specifies the prime factors need not be distinct.


geometry
rectangle
perimeter
quadratics
algebra
quadratic formula

[quote="ZhangPeijin"][quote="crazychinesemaniac"]Really need help on this problem..my head hurts from thinking about it..

There is a playground being built by a daycare center, it is built such that the wall of the building is one of the sides of the playground. The playground is a perfect rectangle with a perimeter of 150 units. The area of the playground is 1500 units^2. The side of the daycare that is part of the perimeter of the playground is one of the long sides of the rectangle. How long is the wall of the daycare that is also part of the playground perimeter?[/quote]
Do you count the wall as part of the 150 units?[/quote]

Yea it counts

[quote="ZhangPeijin"]Well then I think it's not possible.
You have a perimeter of 150.
To get the greatest area, you make a square.
Side lengths of 37.5.
That means the greatest area possible is 1406.25.
Nowhere near your required 1500.[/quote]

umm sry typed it in wrong meant perimeter of 200

[hide]Okay, we have $ lw=1500$ and $ 2l+2w=200$. So $ l+w=100$ and $ w=100-l$. Substituting in the first equation gives
\begin{align*}
l(100-l)&=1500\\
100l-l^2&=1500\\
0&=l^2-100l+1500\\
\end{align*}

Using the quadratic formula, $ l=50 \pm 10\sqrt{10}$. Since the length is longer, $ l=\boxed{50 + 10\sqrt{10}}$[/hide]


algebra
polynomial
complex numbers
Galois Theory
superior algebra
superior algebra solved

Since a couple of days i was asking myself:
Does all subfield of the complex numbers are closed by conjugation?
that is if $\mathcal{K}$ is a subfield of $\mathbb{C}$ then: $z\in\mathcal{K}\Longrightarrow \overline{z}\in\mathcal{K}$?

I'll appreciate any help.

Of course not. Let $r$ be one of the non-real roots of $z^3+z+1=0$. The field of numbers $a+br+cr^2$, where $a,b,c$ are rational, is not closed under conjugation. In particular, it does not contain the conjugate of $r$.
The proof belongs to Galois theory- it is based on the fact that this field has degree 3 over $\mathbb{Q}$, so any automorphism of it must have order dividing 3.
Any polynomial of odd degree, irreducible over $\mathbb{Q}$, with some non-real roots, would allow a similar construction.

Thanks, it was a nice example, well that question have a little history, it became from another about the field of constructible numbers, well it's easy to see that this field is close by radication (for each $z\in\mathcal{K}$ there is a $w\in\mathcal{K}$ such $w^2=z$),  and i was hoping that it is minimal with this property (that is, every subfield of $\mathbb{C}$  closed by radication must have inside the field of constructible numbers) well it's almost true i could prove that this field is minimal among those subfields of the complex plane tha are close by radication an conjugation at the same time.

So it remains the question: does the field of constructible numbers are minimal among those closed by radication?
well i hope that you find this question as interesting as i did.

The intersection of such a field with $\mathbb{R}$ must be closed under square roots of positive numbers. The real constructible numbers are minimal here. Adjoin $i$, which must be in the field, and we get the constructible numbers. Done.

i feel kinda dumb of not noticed that  :P , i had it almost at my nose (this is a spanish expresion...) and i dont figuered it out,  well that was my original problem, but i was trying  to prove that  each subfield closed by radication must be closed by conjugation...

again i'll apprechiate any comment about this......


Let $ A \equal{} \{1,2,\ldots,2008\}$
a) Find the number of subset of $ A$ which satisfy : the product of its elements is divisible by 7
b) Let $ N(i)$ denotes the number of subset of $ A$ which sum of its elements remains $ i$ when divided by 7. Prove that $ N(0) \minus{} N(1) \plus{} N(2) \minus{} N(3) \plus{} N(4) \minus{} N(5) \plus{} N(6)\minus{}N(7) \equal{} 0$

EDITED : thx for cosinator.. BTW, your statement and my correction give 80% hint of the solution :D

[quote="dyd"]b) Let $ N(i)$ denotes the number of subset of $ A$ which sum of its elements remains $ i$ when divided by 7. Prove that $ N(0) \minus{} N(1) \plus{} N(2) \minus{} N(3) \plus{} N(4) \minus{} N(5) \plus{} N(6) \equal{} 0$[/quote]

Are you sure this is the problem statement?  It seems to me it should be $ N(1) \minus{} N(2) \plus{} N(3) \minus{} N(4) \plus{} N(5) \minus{} N(6) \equal{} 0$

I'll solve the (b) part, since part (a) was an easy one.
We're going to proof that $ N(i) \equal{} N(7 \minus{} i)$ for $ i \equal{} 1,2,3$.
Obviously, $ N(0) \equal{} N(7)$
Now, consider a set $ B$, a subset of $ A$, which contains $ b_1,b_2,b_3,...,b_k$
so that:
$ b_1 \plus{} b_2 \plus{} b_3 \plus{} ... \plus{} b_k\equiv i (\mod 7)$
then, there's exactly one solution of $ x_1 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_k\equiv \minus{} i (\mod 7)$
that corresponds with the previous solution, which is:
$ (2009 \minus{} b_1) \plus{} (2009 \minus{} b_2) \plus{} (2009 \minus{} b_3) \plus{} ... \plus{} (2009 \minus{} b_k)\equiv \minus{} i (\mod 7)$
Because there's a bijection, we can conclude that $ N(i) \equal{} N(7 \minus{} i)$
or, equivalently:
$ N(0) \minus{} N(1) \plus{} N(2) \minus{} N(3) \plus{} N(4) \minus{} N(5) \plus{} N(6) \minus{} N(7) \equal{} 0$
[Q.E.D.]

Is this proof right, or wrong?

You can also make the bijection between sets that contain 6 and the ones that don't. If two sets are identical except that one contains 6 and the other doesn't, then one has remainder i and the other has remainder 7-i.

That gives you the result for all n > 5.

The second part of the question is asking about the sum of the elements, not the product.  (Was the first part intended to ask about the sum as well?)

wangsacl, that's right (although it's not the best presentation possible).

Yep. I still don't know how to present that solution, since bijection is a new technique for me. :D


Let $ a,b$ and $ c$ be the sides of right triangle. Prove that $ \left (\cos\frac{\alpha\minus{}\beta}{2}\right )^2 \geq\frac{2ab}{c^2}$.

[hide="Easy"]

c is the hypotenuse? Let's say so.

Let $ \alpha>\beta$
$ \sin\alpha\equal{}\frac{a}{c},\cos\alpha\equal{}\frac{b}{c}$

$ \left (\cos\frac{\alpha\minus{}\beta}{2}\right )^{2}\equal{}\left (\cos\frac{2\alpha\minus{}90^\circ}{2}\right )^{2}\equal{}\left (\cos (\alpha\minus{}45^\circ)\right )^{2}\equal{}\frac12(\sin\alpha\plus{}\cos\alpha)^2\equal{}\frac{(a\plus{}b)^2}{2c^2}$

So we have to prove $ \frac{(a\plus{}b)^2}{2c^2}\ge\frac{2ab}{c^2}$. It's obvious, by AM-GM. [/hide]

hello, substituting $ \beta\equal{}\frac{\pi}{2}\minus{}\alpha$ and applying the addition formulas we get
$ 1\plus{}2\sin(\alpha)\cos(\alpha)\geq\frac{4ab}{c^2}$
and this is equivalent with $ 1\plus{}\frac{2ab}{c^2}\geq\frac{4ab}{c^2}$ and from here we get $ (a\minus{}b)^2\geq0$.
Sonnhard.

[quote="Bugi"][hide="Easy"]

c is the hypotenuse? Let's say so.

Let $ \alpha > \beta$
$ \sin\alpha \equal{} \frac {a}{c},\cos\alpha \equal{} \frac {b}{c}$

$ \left (\cos\frac {\alpha \minus{} \beta}{2}\right )^{2} \equal{} \left (\cos\frac {2\alpha \minus{} 90^\circ}{2}\right )^{2} \equal{} \left (\cos (\alpha \minus{} 45^\circ)\right )^{2} \equal{} \frac12(\sin\alpha \plus{} \cos\alpha)^2 \equal{} \frac {(a \plus{} b)^2}{2c^2}$

So we have to prove $ \frac {(a \plus{} b)^2}{2c^2}\ge\frac {2ab}{c^2}$. It's obvious, by AM-GM. [/hide][/quote]
Yes, its easy. I intended to post this topic in High School Basics, but I've posted it here by accident.


logarithms
function
inequalities
inequalities proposed

a,b,c >= 4n/5 and abc = n^3 with n € IN* , prove that :
[img]http://www.gnux.be/latex/data/404cc42603d99e7faf9577a55774c040.png[/img]

Let $ f(x) \equal{} \frac {1}{\sqrt {n^3 \plus{} 3x^3}} \plus{} \frac {9}{16n\sqrt {n}}\ln x\ \left(x\geq \frac {4}{5}n\right)$,

$ f'(x) \equal{} \frac {1}{16n\sqrt {n}x}\cdot \frac {(n^3 \plus{} 3x^3)^3 \minus{} (4nx^2)^3}{\sqrt {(n^3 \plus{} 3x^3)^3}\{(n^3 \plus{} 3x^3)^2 \plus{} 4nx^2(n^3 \plus{} 3x^3) \plus{} 16n^2x^4)\}}$

$ \equal{} \frac {1}{16n\sqrt {n}x}\cdot \frac {(x \minus{} n)(3x^2 \minus{} nx \minus{} n^2)}{\sqrt {(n^3 \plus{} 3x^3)^3}\{(n^3 \plus{} 3x^3)^2 \plus{} 4nx^2(n^3 \plus{} 3x^3) \plus{} 16n^2x^4)\}}$.

Let $ g(x) \equal{} 3x^2 \minus{} nx \minus{} n^2 \equal{} 3\left(x \minus{} \frac {n}{6}\right)^2 \minus{} \frac {13}{12}n^2$, $ g(x)$ is a concave down qudratic function, thus by $ g\left(\frac {4}{5}n\right) \equal{} \frac {3}{25}n^2 > 0$.
yielding $ g(x) > 0$ for $ x\geq \frac {4}{5}n$, that is to say, $ f(x)$ has the same sign as $ x \minus{} n$.$ \therefore f(x)\geq f(n)\Longrightarrow f(x)\geq \frac {1}{2n\sqrt {n}} \plus{}\frac {9}{16n\sqrt {n}} \ln n\geq \frac {1}{2n\sqrt {n}}$ for $ n \equal{} 1,\ 2,\ \cdots$.

$ \therefore f(a) \plus{} f(b) \plus{} f(c)\geq \frac {3}{2n\sqrt {n}}\Longleftrightarrow \frac {1}{\sqrt {n^3 \plus{} 3a^3}} \plus{} \frac {1}{\sqrt {n^3 \plus{} 3b^3}} \plus{} \frac {1}{\sqrt {n^3 \plus{} 3c^3}}\geq \frac {3}{2n\sqrt {n}}$. Q.E.D.

I have a similar idea but different method !
the inequality is ( after some transformations ) : [img]http://www.gnux.be/latex/data/72744ed705184ced3008175e3b3c08e4.png[/img]
with : [img]http://www.gnux.be/latex/data/785faa0a2ce2fc851b066e6acbf4fec2.png[/img]
now, let's consider : [img]http://www.gnux.be/latex/data/ef4597ba4679d0c6412c6bcaf214ac62.png[/img]
after studiying the variations of f we have : for all x >= 4/5  f(x)>= f(1) = 1/2
and using xyz=1 leads to the result !!


inequalities
geometry
circumcircle
trigonometry
geometry solved

let $O$ be the circumcenter of the $ABC$
and let $A'$ be the intersection of line $AO$ and the circumcircle of triangle $BCO$
now prove that:
$\frac{AA'}{OA'}+\frac{BB'}{OB'}+\frac{CC'}{OC'}\geq 4.5$

A small but important condition has been left out ;) . In fact, the problem should be:

[color=blue][b]Problem.[/b] Let ABC be an [b]acute-angled[/b] triangle with circumcenter O. Let the lines AO, BO, CO meet the circumcircles of triangles BOC, COA, AOB at the points A', B', C', respectively (apart from O). Prove that

$\frac{AA^{\prime}}{OA^{\prime}}+\frac{BB^{\prime}}{OB^{\prime}}+\frac{CC^{\prime}}{OC^{\prime}}\geq\frac92$.[/color]

[i]Solution.[/i] Since the point O is the circumcenter of triangle ABC, we have AO = BO = CO and the central angle theorem yields < COA = 2 < CBA = 2B and < BOC = 2 < BAC = 2A. In the isosceles triangle BOC (isosceles because of BO = CO), the base angle equals $\measuredangle OCB=\measuredangle OBC=\frac{180^{\circ}-\measuredangle BOC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$. Since the points B, O, C, A' lie on one circle, < A'BC = < A'OC. Thus,

< A'BO = < A'BC + < OBC = < A'OC + < OBC = (180° - < COA) + < OBC
= (180° - 2B) + (90° - A) = (180° - A - B) + 90° - B
= C + 90° - B         (since 180° - A - B = C according to the sum of angles in a triangle)
= 90° - (B - C).

On the other hand, since the points B, O, C, A' lie on one circle, < OA'B = < OCB = 90° - A. Thus, by the sine law in triangle BOA', we get

$\frac{BO}{OA^{\prime}}=\frac{\sin\measuredangle OA^{\prime}B}{\sin\measuredangle A^{\prime}BO}=\frac{\sin\left(90^{\circ}-A\right)}{\sin\left(90^{\circ}-\left(B-C\right)\right)}=\frac{\cos A}{\cos\left(B-C\right)}=\frac{2\sin A\cos A}{2\sin A\cos\left(B-C\right)}$.

Since A = 180° - (B + C) by the sum of angles in triangle ABC, we have sin A = sin (B + C), and this becomes

$\frac{BO}{OA^{\prime}}=\frac{2\sin A\cos A}{2\sin\left(B+C\right)\cos\left(B-C\right)}=\frac{\sin\left(2A\right)}{\sin\left(2B\right)+\sin\left(2C\right)}$

(here we used the formula $2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=\sin x+\sin y$, applied to x = 2B and y = 2C). Since BO = AO, this rewrites as $\frac{AO}{OA^{\prime}}=\frac{\sin\left(2A\right)}{\sin\left(2B\right)+\sin\left(2C\right)}$, so that

$\frac{AA^{\prime}}{OA^{\prime}}=\frac{AO+OA^{\prime}}{OA^{\prime}}=\frac{AO}{OA^{\prime}}+1=\frac{\sin\left(2A\right)}{\sin\left(2B\right)+\sin\left(2C\right)}+1=\frac{\sin\left(2A\right)+\sin\left(2B\right)+\sin\left(2C\right)}{\sin\left(2B\right)+\sin\left(2C\right)}$.

Note that, since triangle ABC is acute-angled, its angles A, B, C are all < 90°, so that sin (2A), sin (2B) and sin (2C) are positive. Defining x = sin (2A), y = sin (2B), z = sin (2C), the above equation becomes $\frac{AA^{\prime}}{OA^{\prime}}=\frac{x+y+z}{y+z}$, and similarly we get $\frac{BB^{\prime}}{OB^{\prime}}=\frac{x+y+z}{z+x}$ and $\frac{CC^{\prime}}{OC^{\prime}}=\frac{x+y+z}{x+y}$.

Now,

$\frac{AA^{\prime}}{OA^{\prime}}+\frac{BB^{\prime}}{OB^{\prime}}+\frac{CC^{\prime}}{OC^{\prime}}=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}$
$=\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)$
$=\frac12\cdot\underbrace{\left(\left(y+z\right)+\left(z+x\right)+\left(x+y\right)\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)}_{\geq 9\text{ by Cauchy-Schwarz}}\geq\frac12\cdot 9=\frac92$,

and the problem is solved.

  Darij


Dear Mathlinkers,
let ABC be a triangle, A1, B1, C1 the midpoints of BC, CA, AB,
A', B', C' the points of contact of the A, B, C-excercles with BC, CA, AB,
A2, B2, C2 the midpoints of AA', BB', CC'.
Prove that the lines A1A2, B1B2, C1C2 meet in on point.
Sincerely
Jean-Louis

Dear Jean-Louis,

If $ P$ is a point with cevian triangle $ A'B'C'$, and $ A_{2}$, $ B_{2}$, $ C_{2}$ are the midpoints of the segments $ AA'$, $ BB'$, $ CC'$, respectively, then the lines $ A_{1}A_{2}$, $ B_{1}B_{2}$, $ C_{1}C_{2}$ are concurrent.

Regards,
Cosmin

See also [url=http://mathlinks.ro/Forum/viewtopic.php?t=193918][b][size=100][color=red]Here[/color][/size][/b][/url].    :read:

kostas Vittas.

Dear Cosmin and mathlinkers,
yes, this problem can be generalized like you have presented and proved with the well knowed "cevian nest theorem" which I found in the work od Döttl and for which I found a synthetic proof (look at my site).
Now, I propose you to go further.
What is the geometric nature of this meet point for the problem and your generalization?Sincerely
Jean-Louis


calculus
integration
function
trigonometry
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data[/b]
(From an Integration Table)
Prove,
$ {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx} = {{\frac {1}{2}}{(}{{{x}{\sqrt {{{a}^{2}} - {{x}^{2}}}}} + {{{a}^{2}}{\arcsin{\frac {x}{a}}}}}{)}}{,}{\,}{\,}{\,}{\,}{\,}{\,}{{{|}{x}{|}}{\leq}{{|}{a}{|}}}$

[b]2. Relevant equations[/b]
Knowledge of integration techniques involving the form,
$ {\sqrt {{{a}^{2}} - {{x}^{2}}}}$

[b]3. The attempt at a solution[/b]
[hide="My attempt at a solution,"]
What is bothering me about this integral is that I do not have a $ {x}$ term on the outside of the radical which is preventing from evaluating this integral by normal convention.

In addition, applying trigonometric substitution methods by rewriting the integral as,
$ {\int}{{a}{\sin{\theta}}}{{(}{dx = ?}{)}}$

seems to lead me no where since rewriting $ {dx}$ in terms of $ {\theta}$ just doesn't seem doable.
[/hide]
Any ideas on how to prove this?

Thanks,

-APSStudent

The standard technique taught in most mainstream calculus textbooks is trigonometric substitution.

We would like to write $ x\equal{}a\sin \theta$ (or $ a\cos\theta,$ but it's more convenient in quadrants other than the first quadrant if you use the sine.) Note that this is already a "backwards" substitution. To write it "forward" we would have to write $ \theta\equal{}\arcsin\left(\frac{x}{a}\right).$

Then we can make the change of variable quite directly: $ dx\equal{}a\cos\theta\,d\theta.$ Put that in the integral to get

$ \int\sqrt{a^2\minus{}x^2}\,dx\equal{}\int(a\cos\theta)(a\cos\theta)\,d\theta\equal{} a^2\int\cos^2\theta\,d\theta.$

Now we need a technique for finding the integral of $ \cos^2\theta.$

That's done using the half angle formula for cosine, if I recall right.

Hey,

[quote="t0rajir0u"]
Try integration by parts.
[/quote]
Not exactly sure how to start directly with integration by parts (IBP),
[hide="My attempt directly with integration by parts"]
Recall,
$ {\int}{d{[}{u(x)}{v(x)}{]}} = {\int}{\biggl[}{u(x)}{d[v(x)]} + {v(x)}{d[u(x)]}{\biggl]}$

$ {\int}{u(x)}{d[v(x)]} = {{u(x)}{v(x)}} - {\int}{v(x)}{d[u(x)]}$

So, one of the things that first stumped me was how should I reexpress,
$ {I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}$

As the two separate functions $ {u(x)}$ and $ {d{[}{v(x)}{]}}$ to correctly integrate by parts.

A little hint on the above would be helpful, because honestly I am not sure how (correctly for IBP) to reexpress the above as the two separate functions: $ {u(x)}$ and $ {d{[}{v(x)}{]}}$.
[/hide]
[quote="Kent Merryfield"]
The standard technique taught in most mainstream calculus textbooks is trigonometric substitution.

We would like to write $ x = a\sin \theta$ (or $ a\cos\theta,$ but it's more convenient in quadrants other than the first quadrant if you use the sine.) Note that this is already a "backwards" substitution. To write it "forward" we would have to write $ \theta = \arcsin(\frac {x}{a}).$

Then we can make the change of variable quite directly: $ dx = a\cos\theta\,d\theta.$ Put that in the integral to get

$ \int\sqrt {a^2 - x^2}\,dx = \int(a\cos\theta)(a\cos\theta)\,d\theta = a^2\int\cos^2\theta\,d\theta.$
[/quote]
Thanks for the reply Kent Merryfield.

When applying the "forward" trigonometric substitution I was not sure of the following,
[hide="My attempt at the trigonometric substitution,"]
Let,
$ {I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}$

Consider the right triangle,
[img]http://s2d1.turboimagehost.com/t/327047_Trigonometry.JPG[/img]
So,
$ {{a}{\sin{\theta}}} = {\sqrt {{{a}^{2}} - {{x}^{2}}}}$

Solving by "forward" substitution,
$ {x} = {\pm}{\sqrt {{{a}^{2}}{{\cos}^{2}{\theta}}}}$

However, how would you justify which root to take: the positive one or the negative one?
[/hide]
Thanks,

-APSStudent

Hey,

So any help on the above, like a little clarification on,

Which root to take when solving for $ {x}$ via "forward" substitution,
$ {x} \equal{} {{\pm}{\sqrt{{{a}^{2}}{\cos}^{2}{\theta}}}}$

Thanks,

-APSStudent


function
complex numbers
complex analysis

Suppose $ f$ is an analytic funtion of the complex variable $ z\equal{}x\plus{}iy$ given by
\[ f(z)\equal{}(2x\plus{}3y)\plus{}ig(x,y)\]
where $ g(x,y)$ is a real-valued function of the real variables $ x$ and $ y$. If $ g(2,3)\equal{}1$, what is the value of $ g(7,3)$?

[hide="solution"]The CR equations are
\[ \frac{\partial g}{\partial y}\equal{}2\\
\frac{\partial g}{\partial x}\equal{}\minus{}3\]
which has the general solution $ g(x,y)\equal{}2y\minus{}3x\plus{}c$.  Since $ c\equal{}g(2,3)\equal{}1$ we have $ g(7,3)\equal{}\minus{}14$.[/hide]

Does this work, too?

The function $ h(z) \equal{} (2 \minus{} 3i)z\equal{} (2x \plus{} 3y) \plus{} i(\minus{}3x \plus{} 2y)$ is surely analytic, so $ f(z) \minus{} h(z) \equal{} i(g(x, y) \plus{} 3x \minus{} 2y)$ must be analytic, as well.  However, this function is also pure imaginary, so it must be constant, $ g(x, y) \equal{} \minus{}3x \plus{} 2y \plus{} c$ for some $ c$.  Then we plug in to find $ c$ and plug in again to find the desired value of $ g$.


inequalities
logarithms
function
calculus
calculus computations

With my solution this seems to be a weak inequality, so I may be making a mistake somewhere. Please check.

$\ln\left( \sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n]{n}}}}}}\right) = \ln \prod_{k=2}^{n}k^{\frac{1}{k!}}= \sum_{k=2}^{n}\frac{\ln k}{k!}< \sum_{k=2}^{\infty}\frac{\ln k}{k!}<$
$< \frac{7\ln2}{12}+\frac{\ln 3}{6}+\sum_{k=5}^{\infty}\frac{k}{k!}< \frac{7\cdot694}{12\cdot1000}+\frac{1099}{6\cdot1000}+e-\frac{8}{3}= e-\frac{1559}{750}<$
$< \frac{2719}{1000}-\frac{1559}{750}= \frac{1921}{3000}< \frac{2}{3}< \ln 2$

Therefore $\sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n]{n}}}}}}< 2$

I am a little suspicious of my solution because the other problem with nested radicals you posted is very hard (I have been working on it for maybe 4 hours with only little progress), while this one I solved immediately... although it took nearly 20 minutes to do the numerical work at the end. I could have used worse estimates for $\ln 2$ and $\ln 3$. By the way, is it possible to calculate the series that I did numerical work for?

edited: fixed things

The inequality you ask about comes from summing the first three terms outside the series and then use that $\ln k < k$ in the remaining terms of the series.

a short proof.(but I just prove a lemma,you can try to prove it by the lemma)
Lemma:
$\sqrt[k]{(k+1)2}\le 2$ for $k \ge 3$
the lemma can be proved quickly.
If somebody want to know the detail I will post it.

[quote="Moubinool"]Prove the inequality $\sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n]{n}}}}}}< 2,\,n \geqslant2.$
(A. Kukush)[/quote]
OK,It is my pleasure.
Use the lemma we have
$\sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n]{n}}}}}}< \sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n]{2n}}}}}}$
$<\sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n-1]{2(n-1)}}}}}}$
$<\sqrt{2\sqrt[3]{{3\sqrt[4]{{4\ldots \sqrt[n-2]{2(n-2)}}}}}}$
$<\cdots$
$<\sqrt{2\cdot 2}$
Have you seen it?

ok thats nice.

maybe i should have been clearer but what i meant was to see a proof the the lemma.

[quote="zhaobin"]
Lemma:
$\sqrt[k]{(k+1)2}\le 2$ for $k \ge 3$
the lemma can be proved quickly.
[/quote]

You have to prove $2k+2 \leq 2^{k}$ for $k \ge 3$

just study the function $f(x)=2^{x}-2x-2$ for  $x \ge 3$

the function is increasing on [3;+oo[ with f(3)=0


For positive real numbers a,b,c it is known, a+b+c=1.
Prove that (1+a)(1+b)(1+c)>=8(1-a)(1-b)(1-c)

 :D Challenging  :?:  I think so

Not sure if this is right...

[hide]$1 + a + b + c + ab + bc + ca + abc \ge 8(1 - a - b - c + ab + bc + ca - abc)$
$2 + ab + bc + ca + abc \ge 2ab - 2abc + 2bc - 2abc + 2ca - 2abc + 6ab + 6bc + 6ca - 2abc$
$2 + ab + bc + ca + abc \ge 2ab(1-c) + 2bc(1-a) + 2ca(1-b) + 6ab + 6bc + 6ca - 2abc$
$2 + ab + bc + ca + abc \ge 2ab(a + b) + 2bc(b + c) + 2ca(c + a) + 6ab + 6bc + 6ca - 2abc$
$2 + ab + bc + ca + abc + 4abc \ge 2a^2b + 2ab^2 + 2b^2c + 2bc^2 + 2c^2a + 2ac^2 + 6ab + 6bc + 6ca -2abc + 4abc$
$2 + ab + bc + ca + 5abc \ge 2(a + b)(b + c)(c + a) + 6ab + 6bc + 6ca - 2abc$
$2 + ab + bc + ca + 5abc \ge 6ab + 6bc + 6ca - 4abc$
$2 \ge 5ab + 5bc + 5ca - 9abc$
$2 - 6abc \ge 5(ab + bc + ca - 3abc)$
$2 - 6abc \ge 5(ab(1-c) + bc(1-a) + ca(1-b))$
$2 - 6abc \ge 5(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b)$
$2 + 4abc \ge 5(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc)$
$2 + 4abc \ge -5abc$
$2 \ge -9abc$

Since $a, b, c > 0$, $abc > 0$. Thus, $-9abc$ is negative, making it less than 2.[/hide]

posted before see here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=8%281-a%29%281-b%29%281-c%29&t=45596[/url]

 ;)

Well Thank you.
I really didn't knew it was posted ealier.
 :lol:


Given a triangle $ ABC$ such that $ AB\equal{}2,\ AC\equal{}1$, the line of bisector of $ \angle{A}$ intersects with $ BC$ at $ D$. If $ CD\equal{}\sqrt{6}$, then find the length of $ AD$.

hello kunny, are you sure that this triangle exists?
Sonnhard.

You are right. I found the answer $ AD^2\equal{}\minus{}10$, which is contradicts.
I will make confirmation of the context to my students.
Sorry, bothering you.

kunny


The problem is an attachement. I just named it Quiz # 3 because its corrections for a quiz and he expects you to correct the quiz that you got wrong with hardly any help! If you got it wrong there, how can you do it outside of class!!! Haha. Anyway help would be great. Thanks.

well...first, what have you tried before you came for help...[note this is not really a homework help forum]

Altheman's right, we're not here to give people answers.
It's about the solution:
[hide="hint"]
Try factoring the trinomials.
Then get a gcd.
[/hide]

I got it down to this. (attachment) Now what?


Prove that the polynomial $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$
could not be written as a product of first degree polynomials with real coefficients .Prove it and generalise it .

If $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=l_{1}(x,y,z)l_{2}(x,y,z)$, then the LHS would have "too many roots"(a union of 2 planes.)


probability
geometry
rectangle
number theory
relatively prime

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[hide]
Clearly there are $2^{9}=512$ possible grids.  We use PIE to count the complement of what we want: the number of grids with 2-by-2 red squares.

There are 4 2-by-2 squares in the grid, so we can choose one and then color the other squares any way we want.  So $4*2^{5}=2^{7}=128$ grids.

But we have overcounted.  We subtract off the grids with two 2-by-2 red squares.  There are 6 ways to have 2 2-by-2 red squares.  In four of these there are 3 remaining squares, so $4*2^{3}=2^{5}=32$ grids.  In the other two there are 2 remaining  squares, so $2*2^{2}=2^{3}=8$ grids.  Subtracting off, we get $128-32-8=88$ grids.

Now we have undercounted.  We add back in the grids with three 2-by-2 red squares.  There are 4 ways to have this, and one remaining square, so $4*2^{1}=2^{3}=8$ grids.  Finally we subtract out the one grid where all squares are red.  The total is $88+8-1=95$ grids.

This is the complement, so the number of grids without 2-by-2 red squares is $512-95=417$.  The probability is $\frac{417}{512}$, which is in simplest form, so $m+n=417+512=929$.
[/hide]

[hide]
We can use PIE to count the number of different colorings with a red $2\times2$ square.

Beginning with a red $2\times2$ square in either of the $4$ possible locations, there are $2^{5}$ ways to color the remaining $5$ squares, so a total of $4\times 2^{5}= 2^{7}$ different colorings.

But this double counts the colorings where there is either a red $3\times 2$ rectangle, or $2$ red $2\times 2$ squares sharing a corner square.

There are $4$ possible locations for the $3\times 2$ rectangle, and $2^{3}$ ways to color the remaining squares, so $2^{5}$ total number of colorings.

There are $2$ possible locations for the $2$ $2\times 2$ squares sharing a corner, and $2^{2}$ ways to color the remaining squares, so $2^{3}$ total number of colorings.

The $1$ $3\times 3$ red square needs to be be subtracted once to be counted.

The total number of different colorings with a red $2\times 2$ square is then equal to
\[2^{7}-(2^{5}+2^{3})+2^{3}-1 = 95 \]
The total number of colorings without a red $2\times 2$ square is then $2^{9}-95 = 417$, so the desired probability is
\[\frac{417}{512}, \]
and $m+n = 417+512 = \boxed{929}$.
[/hide]

[hide="Solution"] Draw the following grid:
[asy]
dot((0,0));
dot((1,0));
dot((2,0));
dot((0,1));
dot((0,2));
dot((1,1));
dot((1,2));
dot((2,1));
dot((2,2));
dot((3,0));
dot((3,1));
dot((3,2));
dot((3,3));
dot((0,3));
dot((1,3));
dot((2,3));
draw((0,0)--(3,0)--(3,3)--(0,3)--(0,0));
draw((1,0)--(1,3)--(2,3)--(2,0)--(1,0));
draw((0,2)--(3,2)--(2,2)--(0,2));
draw((0,1)--(3,1));
[/asy]
Let the top left 2 by 2 square be $A$, the top right 2 by 2 square be $B$, the bottom left 2 by 2 square be $C$, and the bottom right 2 by 2 square be $D$.

Let $A_1$ be the set for when $A$ is completely white, $B_1$ be the set for when $B$ is completely white, and so forth.

We desire to find the probability that $A_1, B_1, C_1$ or $D_1$ are completely white.  To do this, we find the probability that $P(A_1\cup B_1\cup C_1\cup D_1)$
Remark that $P(A_1\cup B_1\cup C_1\cup D_1)=4P(A_1)-4P(A_1\cap B_1)-2P(A_1\cap D_1)+4P(A_1\cap B_1\cap C_1)-P(A_1\cap B_1\cap C_1\cap D_1)$ in this case.

Remark that the probability that $A_1$ is completely white is $\frac{1}{16}$, that the probability that $A_1$ and $B_1$ are completely white is $\frac{1}{64}$, the probability that $A_1$ and $D_1$ are completely white is $\frac{1}{128}$, the probability that $A_1, B_1,$ and $C_1$ are completely white is $\frac{1}{256}$, and the probability that all of $A_1, B_1, C_1, D_1$ are completely white is $\frac{1}{512}$.

So hence, our total probability is $4*\frac{1}{16}-4*\frac{1}{64}-2*\frac{1}{128}+4*\frac{1}{256}-\frac{1}{512}=\frac{1}{4}-\frac{1}{16}-\frac{1}{64}+\frac{1}{64}-\frac{1}{512}=\frac{1}{4}-\frac{1}{16}-\frac{1}{512}$.  From here, note that this is equal to $\frac{95}{512}$.  Since this is the probability of failure, we find probability of success by using $1-\text{Failure}$ to give $\frac{417}{512}$ or $\boxed{929}$. [/hide]

I would like to point out that there is a nice way to solve this problem with out the Principle of Inclusion-Exclusion:
We proceed by complementary counting. First, notice that if there is a $2 \times 2$ red square, then the center cell in the $3 \times 3$ grid must be red. Now, of the remaining eight cells, we shall refer to the four non-corner cells as edges. We now do casework on how many red edges there are.
First, observe that there must be at least two red edges in order to have a $2 \times 2$ red square, so we can ignore the cases with zero or one red edges. Now, if there are two red edges, we see that there are $4$ ways to choose which edges these two are (since we cannot have the two red edges be opposite each other in the $3 \times 3$ grid). Once we have selected these two red edges, it is necessary that the corner cell between these two edges be red; meanwhile, the remaining three cells can be any color, so for this case we have a total of $4 \times 2^3 = 32$ possible colorings. Now, suppose that there are three red edges. Then there are $4$ ways to choose which edges these three are, and for each set of three red edges, there are $2^4 - 2^2$ ways to color the remaining four edges (since any coloring will work, except those in which the two corners between two red edges are both colored blue). Therefore, our total for this case is $4 \times \left(2^4 - 2^2\right) = 48.$ Finally, if there are four red edges, then it's not hard to see that $2^4 - 1$ colorings of the remaining four cells yield a $2 \times 2$ red square. Hence, our total for this case is $2^4 - 1 = 15.$ Finally the probability that we seek is then \[1 - \frac{32 + 48 + 15}{2^9} = \boxed{\frac{417}{512}}.\]

This was an AIME problem that was in Intermediate C&P

An elementary application of inclusion exclusion principle kills the problem.

oops I came back to write the sol:

We can proceed by complementry counting and Principle of Inclusion and Exclusion.
$$\frac{512 - (4 \cdot 2^5 - (2 \cdot 2^2 + 4 \cdot 2^3) + 4 \cdot 2 -1)}{512} = \frac{417}{512}.$$
Hence, the answer is $417 + 512 = \boxed{929}.$

[b]Solution:[/b] We proceed with PIE and complementary counting. Let's try to find the probability that a 2 by 2 red squares DOES appear. **NOTE: When I refer to "x" number of 2-by-2 red squares in the casework that follows, I mean the number of *distinct* 2-by-2 squares (they can overlap but they are not the same). If there is at least one such square, we have $4 \cdot 2^5 = 128$ possibilities for the grid. If there is at least two such squares (that can be side by side or diagonally opposite), we have $4 \cdot 2^3 + 2 \cdot 4 = 40$ possibilities. If there are at least three such squares, we have $4 \cdot 2 = 8$ possibilities. And, of course, if there are at least four of these squares there is only $1$ possibility (whole grid is red). Hence, applying PIE, there is a $\frac{128-40+8-1}{2^9} = \frac{95}{512}$ probability that there is a 2-by-2 red square in the grid. Hence our answer is $\frac{512-95}{512} = \frac{417}{512} \Rightarrow \boxed{929}$.


k natural number

m=3*k + 2 can never be a perfect square.

How to prove it?

let, x be a number, so, it is either 0 or 1 or 2(mod3)
in forst case square is also of form '0'  ;in second it of form 1; in third $ 2^2\equal{}4\equal{}1$
so, m can be of form 0 or 1 (mod 3) or of form 3k or 3k+1

[quote="analytic"]let, x be a number, so, it is either 0 or 1 or 2(mod3)
in forst case square is also of form '0'  ;in second it of form 1; in third $ 2^2 \equal{} 4 \equal{} 1$
so, m can be of form 0 or 1 (mod 3) or of form 3k or 3k+1[/quote]

It seems that you want to say that any x^2 mod(3) is equivalent to 1 or 0
and m mod(3)=2
so m could never be a square.

If we know that m could never be square how could we produce a number square free?

Easy...
See we know that a square when divided by 3 leaves remainder 0 or 1, but 3k+2 leaves remainder 2 - which is an obvious contradiction....thus proved it an't be a square.

[quote="analytic"]sorry, i didnt understand y're queston. please explain it further like maybe givivng a instance[/quote]

I'm sorry I did a mistake.
I was trying to find a way to know if any number could be stated as square free using some formula.
But I was wrong (for now).

I am sorry for my bad english.
This is my first problem I solve here in MathLinks!! The problems are still some difficult to me, but I wanna learn.

We have
$ 3k\plus{}2\equal{}z^2$ then
$ z^2\equiv 2 (mod 3)$
But the squares of the remainders modulo 3 are 0 or 1.
So we have a contradiction.
 :oops:

This is an age-old thread, and if the problems are difficult, turn to pre-olympiad, intermediate topics or high-school basics


inequalities
number theory unsolved
number theory

$n\in N$. Prove that $\{n\sqrt{2}\}>\frac{\sqrt{2}}{4n}$

Let $n\sqrt 2 =m+\theta, \ 0<theta<1$, then $\theta=\frac{2n^{2}-m^{2}}{2m+\theta}>\frac{\sqrt 2 }{4n}.$

Your "because" does not imply the result...
Try $m=n=1$, then the inequality is wrong.


[quote="gauss1181"]I don't understand properties of floor functions that well. Can someone help me solve this:

f(x)=x*floor(x*floor(x*floor(x))).

a. For what values of x is f(x)=2001?
b. Prove that there are no solutions for f(x)=2002.[/quote]

[hide="My short (but not very honest) solution"]
$ f(x)$ is decreasing for $ x\in(\minus{}\infty,\minus{}1]$
$ f(x)\equal{}0$ for $ x\in(\minus{}1,1)$
$ f(x)$ is increasing for $ x\in[1,\plus{}\infty)$

It's rather easy to verify that $ \forall x\in(\frac{286}{41},7)$, we have $ f(x)\equal{}286x$
Then, since $ \frac{2001}{286}\in(\frac{286}{41},7)$, we have $ f(\frac{2001}{286})\equal{}2001$
More, since $ \lim_{x\to 7^\minus{}}f(x)\equal{}286\cdot 7\equal{}2002$ and since $ f(7)\equal{}2401$, we have no positive solution for $ f(x)\equal{}2002$

It's also rather easy to verify that $ \forall x\in (\minus{}\frac{304}{46},\minus{}\frac{303}{46})$, we have $ f(x)\equal{}\minus{}304x$ But $ f(\minus{}\frac{303}{46})\equal{}1995.84...$ and $ \lim_{x\to\minus{}\frac{303}{46}^\minus{}}\equal{}\frac{303\cdot 304}{46}\equal{}2002.43...^\plus{}$
So no negative solutions for $ f(x)\equal{}2001$ and $ f(x)\equal{}2002$

[/hide]

[hide="My honest (but not very short) solution"]

$ f(x)\equal{}x[x[x[x]]]$

1) some basic infos :
$ f(x)$ is decreasing for $ x\in(\minus{}\infty,\minus{}1]$
$ f(x)\equal{}0$ for $ x\in(\minus{}1,1)$
$ f(x)$ is increasing for $ x\in[1,\plus{}\infty)$
$ f(n)\equal{}n^4$ $ \forall n\in \mathbb Z$

2) An expression of $ f(x)$ when $ |x|>1$
Let $ p\equal{}[x]$. We have $ x\equal{}p\plus{}\{x\}$ (with $ p\in\mathbb Z$ and $ 0\leq \{x\}< 1)$. We have $ |p|>1$
Let $ q\equal{}[p\{x\}$. We have $ p\{x\]\equal{}q\plus{}r$ (with $ q\in\mathbb Z$ and $ 0\leq r< 1)$. We have either $ 0\leq q<p$, either $ p\leq q<0$.
So $ x\equal{}p\plus{}\frac qp\plus{}\frac{r}{p}$
$ |x|>1$ $ \implies$ $ p^2\plus{}q>0$
Let $ s\equal{}(p^2\plus{}q)r$ with $ 0\leq s<p^2\plus{}q$

So $ x\equal{}p\plus{}\frac qp\plus{}\frac{s}{p(p^2\plus{}q)}$ with $ 0\leq s < p^2\plus{}q$

Then :
$ [x]\equal{}p$
$ x[x]\equal{} p^2\plus{}q\plus{}\frac{s}{p^2\plus{}q}$
$ [x[x]]\equal{}p^2\plus{}q$ (since $ 0\leq \frac{s}{p^2\plus{}q}<1$ and
$ x[x[x]]\equal{}p^3\plus{}2pq\plus{}\frac{s\plus{}q^2}{p}$
$ [x[x[x]]]\equal{}p^3\plus{}2pq\plus{}[\frac{s\plus{}q^2}{p}]$ and $ f(x)\equal{}x(p^3\plus{}2pq\plus{}[\frac{s\plus{}q^2}{p}])$

So :
Any $ x$ such that $ |x|>1$ may be written in an unique manner as $ x\equal{}p\plus{}\frac qp\plus{}\frac{s}{p(p^2\plus{}q)}$ with
$ p\in\mathbb Z$, either $ 0\leq q<p$, either $ p\leq q<0$ and $ 0\leq s < p^2\plus{}q$

Then $ f(x)\equal{}x(p^3\plus{}2pq\plus{}[\frac{s\plus{}q^2}{p}])$

3) Discontinuity points of $ f(x)$ when $ |x|>1$
So the discontinuity points of $ f(x)$ are all changes in, $ p,q$ plus the cases where $ s\plus{}q^2\equal{}kp$

The first (changes in $ p$ and $ q$) are $ x\equal{}p\plus{}\frac qp$ and $ f(x)\equal{}x(p^3\plus{}2pq\plus{}[\frac{q^2}{p}])$

The others ($ s\plus{}q^2\equal{}kp$) are :  $ x\equal{}p\plus{}\frac qp\plus{}\frac{kp\minus{}q^2}{p(p^2\plus{}q)}$ $ \equal{}\frac{p^3\plus{}2pq\plus{}k}{p^2\plus{}q}$ with $ k\in[\frac{q^2}{p},\frac{q^2}{p}\plus{}p\plus{}\frac qp)$ (maybe in the reverse order if $ p<0)$
For such values, $ [x[x[x]]]\equal{}p^3\plus{}2pq\plus{}[\frac{s\plus{}q^2}{p}]$ $ \equal{}p^3\plus{}2pq\plus{}k$ and so $ f(x)\equal{}x(p^3\plus{}2pq\plus{}k)$

So : discontinuity points of $ f(x)$ are $ \minus{}1, 1$ and :
$ x_{pq}\equal{}\frac{p^2\plus{}q}{p}$ with $ f(x)\equal{}x(p^3\plus{}2pq\plus{}[\frac{q^2}{p}])$
$ x_{pqk}\equal{}\frac{p^3\plus{}2pq\plus{}k}{p^2\plus{}q}$ with $ f(x_{pqk})\equal{}\frac{(p^3\plus{}2pq\plus{}k)^2}{p^2\plus{}q}$ ($ \equal{}x^2(p^2\plus{}q)$)

4) How to solve $ f(x)\equal{}a$
The problem is just to find two subsequents discontinuity points $ x_i<x_j$ such that $ a\in[f(x_i),f(x_j)]$ (notice that there are two such couples : one positive, one negative). Then :
Compute $ \frac{f(x)}{x}$ for $ x\equal{}x_i$, for $ x\equal{}x_j$ and for $ x\in(x_i,x_j)$ and take the conclusion.

5) Application : Solve $ f(x)\equal{}2001$ and $ f(x)\equal{}2002$
5.1) When $ x>1$
$ f(6)\equal{}1296< 2001 < f(7)\equal{}2401$ $ \implies$ $ p\equal{}6$
A quick computation gives then $ f(6\plus{}\frac 56)$ $ \equal{}x(p^3\plus{}2pq\plus{}[\frac{q^2}{p}])$ $ \equal{}(6\plus{}\frac 56)(216\plus{}60\plus{}4)\equal{}1913.3333...$
So $ q\equal{}5$
Between $ 6\plus{}\frac 56$ and $ 6$, discontinuity points are $ x_{65k}\equal{}\frac{p^3\plus{}2pq\plus{}k}{p^2\plus{}q}$ $ \equal{}\frac{6^3\plus{}60\plus{}k}{41}$ $ \equal{}\frac{276\plus{}k}{41}$ with $ k\in[\frac {q^2}{p},\frac{q^2}{p}\plus{}p\plus{}\frac qp)$ $ \equal{}[\frac{25}{6},\frac{25}{6}\plus{}6\plus{}\frac 56)$

So $ x_{65k}\equal{}\frac{276\plus{}k}{41}$ with $ k\in \{5,6,7,8,9,10\}$ and $ f(x_{65k})\equal{}\frac{(276\plus{}k)^2}{41}$
This gives $ k\equal{}10$ and $ f(\frac{286}{41})\equal{}\frac{286^2}{41}\equal{}1995.02...$

Then the two subsequent discontinuity points we are interested in are :
$ a\equal{}\frac{286}{41}$ with $ f(a)\equal{}\frac{286^2}{41}\equal{}1995.02...$
$ b\equal{}7$ with $ f(b)\equal{}2401$
And we have $ f(x)\equal{}x\frac{f(a)}{a}\equal{}286x$ $ \forall x\in (a,b)$

So $ \frac{2001}{286}\in (a,b)$ but $ \frac{2002}{286}\equal{}7\notin (a,b)$ and so :

$ f(\frac{2001}{286})\equal{}2001$ and no positive solution for $ f(x)\equal{}2002$

5.2) When $ x<\minus{}1$
$ f(\minus{}6)\equal{}1296< 2001 < f(\minus{}7)\equal{}2401$ $ \implies$ $ p\equal{}\minus{}7$
$ q\equal{}\minus{}2$ gives $ f(\minus{}7\plus{}\frac 27)\equal{}x(p^3\plus{}2pq\plus{}[\frac{q^2}{p}])$ $ \equal{}(\minus{}7\plus{}\frac 27)(\minus{}343\plus{}28\minus{}1)\equal{}2221.71...$
$ q\equal{}\minus{}3$ gives $ f(\minus{}7\plus{}\frac 37)\equal{}x(p^3\plus{}2pq\plus{}[\frac{q^2}{p}])$ $ \equal{}(\minus{}7\plus{}\frac 37)(\minus{}343\plus{}42\minus{}2)\equal{}1991.14...$
So $ q\equal{}\minus{}3$

Between $ \minus{}7\plus{}\frac 27$ and $ \minus{}7\plus{}\frac 37$, discontinuity points are $ x_{\minus{}7,\minus{}3,k}\equal{}\frac{p^3\plus{}2pq\plus{}k}{p^2\plus{}q}$ $ \equal{}\frac{k\minus{}301}{46}$  with $ k\in[\frac {q^2}{p},\frac{q^2}{p}\plus{}p\plus{}\frac qp)$ $ \equal{}[\minus{}\frac{9}{7},\minus{}\frac{9}{7}\minus{}7\plus{}\frac 37)$

So $ x_{\minus{}7,\minus{}3,k}\equal{}\frac{k\minus{}301}{46}$ with $ k\in \{\minus{}7,\minus{}6,\minus{}5,\minus{}4,\minus{}3,\minus{}2\}$ and $ f(x_{\minus{}7,\minus{}3,k})\equal{}\frac{(k\minus{}301)^2}{46}$

$ k\equal{}\minus{}3$ $ \implies$ $ f(x_{\minus{}7,\minus{}3,\minus{}3})\equal{}\frac{(k\minus{}301)^2}{46}$ $ \equal{}\frac{(\minus{}304)^2}{46}\equal{}2009.04...$
$ k\equal{}\minus{}2$ $ \implies$ $ f(x_{\minus{}7,\minus{}3,\minus{}2})\equal{}\frac{(k\minus{}301)^2}{46}$ $ \equal{}\frac{(\minus{}303)^2}{46}\equal{}1995.84...$

Then the two subsequent discontinuity points we are interested in are :
$ a\equal{}\frac{\minus{}304}{46}$ with $ f(a)\equal{}\frac{304^2}{46}\equal{}2009.04...$
$ b\equal{}\frac{\minus{}303}{46}$ with $ f(b)\equal{}\frac{303^2}{46}\equal{}1995.84...$

And we have $ f(x)\equal{}\minus{}304x$ $ \forall x\in (a,b)$ and so $ f(x)\in(\frac{303\cdot 304}{46},\frac{304^2}{46})$

And since $ 2001\notin (\frac{303\cdot 304}{46},\frac{304^2}{46})$ and $ 2002\notin (\frac{303\cdot 304}{46},\frac{304^2}{46})$ :

No negative solution for $ f(x)\equal{}2001$ and $ f(x)\equal{}2002$
[/hide]


The arithmetic mean of the nine numbers in the set $ \{9,99,999,9999,...,999999999\}$ is a $ 9$-digit number $ M$, all of whose digits are distinct. The number $ M$ does not contain the digit

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

[hide]We factor as $ 9(1\plus{}11\plus{}...\plus{}111111111)$. Since there are only 9 numbers, there is no 0,A.[/hide]


logarithms
real analysis
real analysis unsolved

yello  :D

show

$ \sum_{k\equal{}1}^\infty\;\dfrac{(\minus{}1)^{k\plus{}1}\;\ln\,(2k\plus{}1)}{2k\plus{}1}\;\equal{}\;\boxed{\dfrac{\pi}{4}\left(\gamma\plus{}\ln\,\left(\dfrac{4\;\pi^3}{\Gamma^4\,\left(\dfrac{1}{4}\right)}\right)\right)}$

You asked this question [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=194624]here[/url] before.


group theory
abstract algebra
invariant
geometry
geometric transformation
rotation
superior algebra

As title , classify finite subgroup of GL_2(R).

Thanks!!

Each finite group $ G\leq GL_n(\mathbb{R})$ induces a $ G$-invariant scalarproduct on $ \mathbb{R}^n$, so is conjugated to a subgroup of $ O_n(\mathbb{R})$. Now $ O_2(\mathbb{R})\equal{}S^1\ltimes\{\pm 1\}$, so each subgroup decomposes into a semidirect of a subgroup of $ S^1$ and a subgroup of $ \{\pm 1\}$. The only finite subgroups of $ S^1$ are the cyclic group generated by a rotation by the $ \tfrac{2\pi}{m}$-angle. Hence the only finite subgroups of $ GL_2(\mathbb{R})$ (up to conjugation) are cyclic and dihedral groups. All cyclic and dihedral group are present.


ratio
geometry
perpendicular bisector
power of a point

In semicircle center O diameter AB radius 4, C is on OB such that OC=2. The perpendicular to AB through O meets the semicircle at D. The perpendicular to AB through C meets the semicircle at E. OD intersects AE at F. The midpoint of EF is G and the perpendicular bisector of EF meets the semicircle at H. Find GH.

[hide="Solution"]
First, to find $ CE$, draw $ OE$ and use Pythagorean. It is now easy to angle chase and find all of the $ 30^\circ\minus{}60^\circ\minus{}90^\circ$ triangles, and to continue by side bashing. Then, draw $ \overline{OI}$, and by the side ratios note that it is parallel to $ \overline{AE}$. Thus, $ m\angle OIH\equal{}90^\circ$, so if we complete the circle and extend $ \overline{HJ}$ to its intersection with the circle, we find that the length of that chord is $ 2(x\plus{}2)$. Thus, by Power of a Point, $ x(2\plus{}(x\plus{}2))\equal{}x^2\plus{}4x\equal{}\frac2{\sqrt3}\cdot\frac{2\plus{}8}{\sqrt3}\equal{}\frac{20}3\implies3x^2\plus{}12x\minus{}20\equal{}0$
$ \implies x\equal{}\frac{\minus{}12\pm\sqrt{384}}6\equal{}\frac{\minus{}12\pm8\sqrt6}6,\,x>0\implies x\equal{}\boxed{\frac{4\sqrt6}3\minus{}2}$.
[/hide]

Correct! Thank you for responding to all of my problems!  :lol: BTW, Pythag on $ \triangle OIH$ also finishes the problem.


induction
function
modular arithmetic
strong induction
combinatorics unsolved
combinatorics

An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a [i]binary sequence of length [/i]$n$.  Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order.  Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.

Might as well post the solution then, though SamE's hint does rather give it away...

[hide="Solution"]
For each of the $a_{n}$ $n$ term sequences $A$ with no $0 1 0$ in it, we can define two $n+1$ term sequences, one beginning with a $1$ and one with a $0$ as follows:
The $k+1$th member of the sequence be the sum of the the $k$th member of the sequence and the $k$th member of the sequence $A$ (modulo 2).
Both of these sequences (the one starting 1 and the one starting 0) do not have $1100$ or $0011$ in them: furthermore, when we vary the sequence $A$ through all $a_{n}$ possible sequences we must get all the possibilities for the $n+1$ member sequence, and nothing can be double-counted.

Hence $b_{n+1}=2a_{n}$ [/hide]

I did it by induction.

[hide="Solution"]Let $A_n$ be the binary sequences of length $n$ containing no three consecutive terms equal to $0$, $1$, $0$ in that order. Let $B_n$ be the binary sequences of length $n$ that contain no four consecutive terms equal to $0, 0, 1, 1$ or $1, 1, 0, 0$ in that order.

We will use strong induction to show that $b_{n} = 2a_{n-1}$ for all positive integers $n$.

Base cases: $n=1, 2, 3$.
$n=1, 2$ are trivial. Since the sequences are too short for any restrictions to apply, $b_{n+1}=2a_n$ because there are two choices ($0$ or $1$) to tack onto the $A$-sequence.
There are $2^3=8$ possible binary sequences of length $3$, so $a_3=7$ since only one of these is eliminated. There are $2^4=16$ possible binary sequences of length $4$, so $b_4=14$ since only two of these are eliminated. Thus, $b_4=2a_3$.

Now, assume that $b_{n+1}=a_{n}$ for all positive integers through $n$.

We now need to prove that $b_{n+2} = 2a_{n+1}$:
We define a series of functions:
$P(n)$ is the number of $n$-term $A$-sequences that end in $0, 1$.
$Q(n)$ is the number of $n$-term $B$-sequences that end in $0, 0, 1$ or $1, 1, 0$.
$X(n)$ is the number of $n$-term $B$-sequences that end in $0, 0, 1$.
$Y(n)$ is the number of $n$-term $B$-sequences that end in $1, 1, 0$.
$f(n)$ is the number of $n$-term $A$-sequences that end in $0$.
$g(n)$ is the number of $n$-term $B$-sequences that end in $0, 0$ or $1, 1$.
$R(n)$ is the number of $n$-term $B$-sequences that end in $0, 0$.
$S(n)$ is the number of $n$-term $B$-sequences that end in $1, 1$.

$a_{n+1}=2a_n-P(n)$ since we can add $0$ or $1$ to $A_n$ for any $A_n$ with the exception that we can’t add $0$ to $A_n$ if it ends in $0, 1$.
Similarly, $b_{n+2}=2b_{n+1}-X(n+1)-Y(n+1)$ since we can add $0$ or $1$ to $B_{n+1}$ for any $B_{n+1}$ with the exception that we can’t add $0$ to $B_{n+1}$ if it ends in $1, 1, 0$ or $1$ to $B_{n+1}$ if it ends in $0, 0, 1$. Thus, $b_{n+2}=2b_{n+1}-Q(n+1) =4a_n-Q(n+1)$ by the induction hypothesis.

So if $b_{n+2} = 2a_{n+1}$, $Q(n+1)=2P(n)$.
$a_n-P(n)=2a_{n-1}-f(n-1)$ since we can get an $A_n$ that does not end in $0, 1$ by adding $0$ or $1$ to $A_{n-1}$ with the exception that we can’t add $1$ to $A_{n-1}$ if $A_{n-1}$ ends in $0$.
Simlarly, $b_{n+1}-X(n+1)-Y(n+1)=2b_n-R(n)-S(n)$ since we can get a $B_{n+1}$ that does not end in $0, 0, 1$ or $1, 1, 0$  by adding $0$ or $1$ to $B_n$ with the exception that we can’t add $1$ to $B_n$ if $B_n$ ends in $0, 0$ and we can’t add $0$ to $B_n$ if $B_n$ ends in $1, 1$. Thus, $b_{n+1}-Q(n+1)=2b_n-g(n)$.
By the induction hypothesis, $b_{n+1}=2a_n$ and $b_n=2a_{n-1}$. Thus, $Q(n+1)=2P(n)$ and therefore $b_{n+2} = 2a_{n+1}$ if and only if $g(n)=2f(n-1)$.

Thus, if we can prove that $g(n)=2f(n-1)$, we are done:
$f(n-1)=2a_{n-3}-f(n-3)$ since if $A_{n-3}$ ends in $0$, $x_{n-2} \neq 1$. Otherwise, we have $2$ choices for $x_{n-2}$.

$R(n)=b_{n-2}-S(n-2)$ since $B_n$ can only end in $0, 0$ if $B_{n-2}$ does not end in $1, 1$.
Similarly, $S(n)=b_{n-2}-R(n-2)$ since $B_n$ can only end in $1, 1$ if $B_{n-2}$ does not end in $0, 0$.
Adding these two equations, we obtain $g(n)=2b_{n-2}-g(n-2)$. By the induction hypothesis, $g(n-2)=2f(n-3)$ and $b_{n-2}=2a_{n-3}$, thus, $g(n)=4a_{n-3}-2f(n-3)=2f(n-1)$.

$Q.E.D.$[/hide]

From a sequence $(x_1,x_2,...,x_n)$ construct another sequence $(y_0,y_1,...,y_n)$ of length $n+1$ such that
$y_0=0$ and $y_i=\sum_{k=1}^{i}{x_k} \pmod{2}$ .It is easy to see that such a constuction is bijective.Also the sequence of the first type has a subsequence $(0,1,0) \Leftrightarrow$ the sequence of the second type has a subsquence $(0,0,1,1)$ or $(1,1,0,0)$.The case when $y_1=1$ can be seen analogously by interchanging the $0$'s ans $1$'s.We are through.....

The less intelligent but quite straightforward recursion solution: Let $c_n$ denote the number of strings counted by $a_n$ that end with $01$, and let $d_n$ denote the number of strings counted by $b_n$ that end with either $001$ or $110$. Verify for small $n$ that $b_{n+1} = 2a_n$ and $d_{n+1} = 2c_n$.

We now establish recursions for $a_n, b_n, c_n, d_n$. To count $a_n$, we have $a_{n-1}$ choices for the first $n-1$ bits and $2$ ways to append an $n$th bit, but we cannot append a $0$ to the $c_{n-1}$ strings ending with $01$. So, we have
\[a_n = 2a_{n-1} - c_{n-1}.\]Count $b_n$ similarly. For each string counted by $d_{n-1}$, there is one way to illegally append $0$ or $1$, so we have
\[b_n = 2b_{n-1} - d_{n-1}.\]To count $c_n$, we have $a_{n-2}$ choices for the first $n-2$ bits and must append $01$ to each of them, but those $(n-2)$-length strings ending with $01$ fail. So, we have
\[c_n = a_{n-2} - c_{n-2}.\]Count $d_n$ similarly. We have $d_{n-2}$ choices for the first $n-2$ bits, and the last bit of the $(n-2)$-length string determines the necessary $(n-1)$th and $n$th bits we must append. However, this fails when the $(n-2)$-length string ends in either $001$ or $110$ since we are forced to append $10$ and $01$, respectively, which is illegal. So, we have
\[d_n = b_{n-2} - d_{n-2}.\]Now apply induction using these recursions to finish.

Consider the map from sequences of the latter form to sequences of the first form by \[ (y_1, \dots, y_{n+1}) \mapsto (y_1 + y_2, y_2 + y_3, \dots, y_n + y_{n+1}). \] It is $2$-to-$1$. The end.

[quote=Williamgolly]The set of $3$ consecutive terms for a sequence $a_n$ are $000, 100, 101, 110, 111, 001, 011.$
The set of $4$ consecutive terms for a sequence $b_n$ are $0000, 0001, 0010, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1101, 1110, 1111.$
In the sequence $b_{n+1},$ there are $n-2$ strings of $4$ consecutive terms.  In a sequence $a_n,$ there are $n-2$ string of $3$ consecutive terms.  Then, for each starting $3$ string in sequence $a_n$, can append 2 different starting number (0 or 1) to the start to make a valid string of $b_{n+1}$. This process can be reversed.  Therefore, we have formed a 1 to 2 bijection between $b_{n+1}$ to $a_n$, so $b_{n+1} = 2a_n$ for all $n$.[/quote]

What if, say, the starting string of one of the terms in $a_n$ is $011$. Then we can't append 0 at the start to make a valid term for $b_n$. How is the bijection valid? Is there anything that I'm missing here?

$a_n$ is the sequence of absolute differences of $b_{n+1}$, where we can pick the first term to be either $1$ or $0$. Thus $b_{n+1} = 2a_n$.

[quote=SamE][hide="Hint"]Take successive differences of a $n+1$-term sequence.[/hide] I'll post the full solution when I find time.[/quote]

legend says he still cannot find time

@above I think I know why @SamE is still not being able to find time. This quote explains the reason. Some great person once said - ``$\textbf{If you cannot prove a statement using induction,}$
$\textbf{try inducting on the inductive step of that induction and if that still doesn't work,}$
$\textbf{try inducting on the inductive step of the inductive step of the induction and so on....}$".

AIME style recursion vibes :ninja: . We proceed using induction. Base case clearly works.

Note that you can perform a map by switching the $0$s with $1$s and vice versa. Note that this map is an involution and thus a bijection. So we can just consider $\dfrac{b_n}{2}$ and then we would only have to handle the condition where no $0$, $0$, $1$, $1$ sequence is there. Then we have to prove that $b_{n+1}=a_n$. $x_n$ denote the number of $n$ lengthed sequence which end with a $0$, $1$ and satisfy the condition for $a_n$. This gives $a_n=2a_{n-1}-x_{n-1}$. Similarly, $y_n$ denote the number of $n$ lengthed sequences which end with $0$, $0$, $1$ and satisfy the condition for $b_n$. This gives $b_n=2b_{n-1} -y_{n-1}$. Thus it suffices to prove that $x_{n-1}=y_{n}$. A bit of case checking gives $x_n=a_{n-2}-x_{n-2}$ and $y_n=y_{n-3}$. Now we do even more induct lol. $x_{n-1}=y_{n}\iff a_{n-3}-x_{n-3}=y_{n-3}\iff a_{n-3}=x_{n-3}+x_{n-4}$ where the last equality follows due to induction. It just leaves us to prove $a_n=x_n+x_{n-1}$. Andddd, guess what? Even more induct! YAYY! :rotfl: We now have $a_n=x_n+x_{n-1}\iff 2a_{n-1}-x_{n-1}=x_n+x_{n-1}\iff x_n=2x_{n-2}$ where the last equality again follows due to induction. This simply follows from that the digit after $x_n$ cannot be $0$ and must be $1$ due to the condition of $a_n$ and then there are $2$ choices for the last digit and we are done.


Is there a generic method to find the summation of the series X=1 to n ∑ x power(m)? I have found a method. But that isn't straight forward. The proof for m = 1 in the above expression is fairly simple and is obtained by representing each item in the summation in terms of 'm' and by taking 'm' as common from the expression and representing the remaining expression as the summation expression itself and obtaining the value. But for higher powers, the method is round about. But there could be some straight forward methods. If any one could share it, it will be great[/hide]

Do you mean $ \sum_{x\equal{}1}^{n}x^m$?

If so, there is [url]http://en.wikipedia.org/wiki/Faulhaber%27s_formula[/url].

Oh! Yes. Exactly.

Thanks a lot. How do you represent these mathematical symbols in this forum. I don' tknow this. Because of this I write lot of texts to explain a simple mathematical expression.


the real root of $ x^3\minus{}15x\minus{}4\equal{}0$ 1s

 $ \sqrt [3]{2 \plus{} 11i} \plus{} \sqrt [3]{2 \minus{} 11i}$

however we can see that the real root is $ 4$ expline

[quote="matrex"]the real root of $ x^3 \minus{} 15x \minus{} 4 \equal{} 0$ 1s

 $ \sqrt [3]{2 \plus{} 11i} \plus{} \sqrt [3]{2 \minus{} 11i}$

however we can see that the real root is $ 4$ expline[/quote]

[u]First remark [/u]: $ x^3 \minus{} 15x \minus{} 4$ has exactly three real roots.

[u]Second remark [/u]: what is your choice for definition of cubic root of a complex number ?
In my knowledge, the equation $ z^3 \equal{} a$ has $ 3$ roots, so $ \sqrt [3]{2 \plus{} 11i} \plus{} \sqrt [3]{2 \minus{} 11i}$ may take 6 values, depending on your choices.

One of these 6 values is, indeed, 4 (and the 5 others are different).

So, could you kindly explain us :
1) what is your definition of  $ \sqrt [3]{2 \plus{} 11i} \plus{} \sqrt [3]{2 \minus{} 11i}$ ?
2) what are you asking exactly ?


Does anyone going to Mathcamp this year know how to play bridge?

Yes! :)

Apparently the above alone is not enough to constitute a non-spam message. So I have added these sentences.

Yes I know how to play bridge.

$\frac{\delta^{2}}{\delta y \delta x}f(x,y) = \frac{\delta^{2}}{\delta x \delta y}f(x,y)$

I can play bridge, but I'm not great at it.

Bridge, poker, and other card games are quite popular at Mathcamp. I'm sure you'll be able to find partners/games.  I think a tournament was organized three years ago, and you are free to organize another one this year!

Don't forget Mao!  :lol:

I know how to play bridge, but I haven't played in a really long time.

[quote="fourierseries"]Bridge, poker, and other card games are quite popular at Mathcamp. I'm sure you'll be able to find partners/games.  I think a tournament was organized three years ago, and you are free to organize another one this year![/quote]

I have a poker set. Should I bring it to camp, or will it be there already? Also, what about other games, like chess et al?

Lastly, we should definitely organize a giant Durak tournament (those that are Russian should know what I'm talking about).

Sure, bring it.  The more games the better.  Chess is another popular one, although last year it was superseded by bughouse.  I'm not Russian and don't know how to play Durak, but I'd like to learn!

OK, sure, I'll bring it.

As for Durak, It would be truly fun if we can get 12 or more people and have a single or even double elimination tournament. Otherwise, it would be a good pastime, but I doubt it's the kind of game that could be considered ``Mathcamp-y'', and so will likely not be very popular. But, then again, who knows?

Bridge was a bit less popular in 2006.  In 2004 and 2005 there were always various people playing in different lounges, and there were tournaments (both for beginners and non-beginners in 2004, I believe).  Many people learned to play at camp.

I am definitely up for Bridge. If anyone wants to play Bughouse, I like to play that a lot too. I can also play Hearts and normal Chess.

Also, is anyone up for sports, like soccer or frisbee?

I have never played Bughouse, but I have heard and read about it a lot and would love to play. I also play normal Chess, but am not very good.

As for sports, I am definitely up for Frisbee, and maybe soccer as well.

If you want to play bridge, you will be able to find others to play with you. That is the way of things.

As for Durak, if you think it might not be considered "Mathcamp-y", then you will be surprised.  There is a perfect sentence to put here, but what it states is one of those things better found out on one's own.

Don't worry about finding people to play Bughouse with.  Last year there was a game going on in the lounge virtually 24 hours a day.

Hearts, soccer and frisbee all sound good.  Also, does anyone play Canasta?  I recently learned how to play from my grandmother and would love to play some at Mathcamp.

That's not true. I woke up really early a few times last year and I was sad cuz nobody else was awake yet :( And people like to sleep at 2am (weekends of course - weekdays we get kicked out at midnight :()

Games start around breakfast time and take breaks for good classes :)

On another note: 3.5 more days (or am I crazy?)

could anybody bring risk? that's like my favourite board game ever.

Settlers of Catan and Diplomacy should also be brought.

Risk, Settlers of Catan, and Diplomacy were all present last year.  I know Risk will be there this year, and the person who brought Settlers of Catan is coming back, so I'm sure he'll bring it again.  Diplomacy I'm not sure about, but most likely it will be there.

Brian, you're right, I exaggerated.  But in general, if at least 20 people were awake and not in class, 4 of them were playing Bughouse.

Since people are discussing games here, does anyone play tennis? I'd love to get to play tennis while I'm at camp.

[quote]Risk, Settlers of Catan, and Diplomacy were all present last year. I know Risk will be there this year, and the person who brought Settlers of Catan is coming back, so I'm sure he'll bring it again. Diplomacy I'm not sure about, but most likely it will be there. [/quote]

We've got a diplomacy set coming, so Risk, Catan, and Diplo should all be covered.  However, you might want to bring your own sets anyway, so that you don't have to keep borrowing a set from the same person.

I am pretty much in agreeance here. Bringing games is good. If there is a game you want to play, and there are materials for it, you will easily find someone to play it with you. Also, you may learn of many fun, new games like Ricochet Robots!


modular arithmetic
algebra
polynomial
number theory

Let $ p(x)\equal{}\sum_{k\equal{}0}^6 x^k$. Find the remainder when $ p(x^7)$ is divided by $ p(x)$.

I feel like I'm missing something obvious - the only solution I've come up with is essentially brute force.

[quote="Mathnerdmo"][hide="I believe this works"]
Note that this is equivalent to asking what $ p(x^7)$ is modulo $ p(x)$.

$ p(x) \equal{} \sum_{k \equal{} 0}^6 x^k \equal{} \frac {1 \minus{} x^7}{1 \minus{} x} \equiv 0 \pmod {p(x)}$ gives $ x^7 \equiv 1$

$ p(x^7) \equiv p(1) \equal{} 7.$
[/hide][/quote]I don't believe that's correct.

What is one supposed to say to that other than, "Your belief is wrong."?

The statement that $ p(x^7) \equiv 7 \bmod p(x)$ can be equivalently phrased as $ p(x) | p(x^7) \minus{} 7$, which one checks as follows:

$ p(x^7) \minus{} 7 \equal{} \sum_{k\equal{}0}^{6} (x^{7k} \minus{} 1)$

and if I cannot convince you that $ p(x) | x^7 \minus{} 1 | x^{7k} \minus{} 1$ then I am not sure what to do.

In any case, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=125861]polynomial modular arithmetic[/url] is what I consider an underused tool; once you're comfortable with modular arithmetic, working with polynomials isn't much harder and makes a lot of otherwise difficult-looking problems automatic.

$ \frac{p(x^7)\minus{}7}{p(x)}\equal{}\minus{}6\plus{}6 x\minus{}5 x^7\plus{}5 x^8\minus{}4 x^{14}\plus{}4 x^{15}\minus{}3 x^{21}\plus{}3 x^{22}\minus{}2 x^{28}\plus{}2 x^{29}\minus{}x^{35}\plus{}x^{36}$

Observe the pattern in the coefficients, and verify that it works.