logarithms
calculus
calculus computations

ln(1+5/1!+25/2!+125/3!+625/4! +...)

I need to find the exact answer of the infinite sum of the series.

Trivial by use Power Series of $e^{x}$. http://en.wikipedia.org/wiki/E%5Ex

@I not see sylow_theory's post  :oops:

Use the fact that:

$e^{x}= \sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

[hide="My Answer"]
$\ln \sum_{n=0}^{\infty}\frac{5^{n}}{n!}= \ln{e^{5}}= 5$[/hide]

EDIT: WOW!  I can't believe TWO posts were made while I was tiyping up this response and editing it.


Let $ a, b, c$ be nonnegative real numbers. Find the best constant $ k$ for this inequality is trues  :) :
\[ (a \plus{} b \plus{} c)^5 \ge k(a \minus{} b)(b \minus{} c)(c \minus{} a)(a^2 \plus{} b^2 \plus{} c^2)\]

[quote="nguoivn"]How about the proof, my friend ? We can solve it by AM-GM  :)[/quote]
I guess that xzlbq got this by employing a computer program.

[size=150]yes,I get it in Bottema program,
http://old.irgoc.org/bbs/dispbbs.asp?boardid=11&Id=443
BQ[/size]


geometry
geometric transformation
rotation
cyclic quadrilateral
geometry solved

In the cyclic quadrilateral $ABCD$, the diagonal $AC$ bisects the angle $DAB$. The side $AD$ is extended beyond $D$ to a point $E$. Show that $CE=CA$ if and only if $DE=AB$.

Since $ABCD$ is cyclic, $\angle ABC=\angle EDC$. We also have $BC=DC$ because $\angle BAC=\angle DAC$. If $DE=BA$ then $\triangle ABC\cong\triangle EDC\Rightarrow CA=CE$. If $CA=CE$ then it is also easy to check that $\triangle ABC\cong\triangle EDC$. (Although $SSA$ is not a criterion for two triangles to be congruent, we can prove that the counter case is impossible by considering the angles.)

$CA=CE \Rightarrow \angle{EAC}=\angle{E} \Rightarrow \angle{E}=\angle{BAC} \Rightarrow \triangle{EDC}\cong\triangle{ABC}$


analytic geometry
function
calculus
integration
modular arithmetic
abstract algebra
algebra unsolved

Let $G$ denote the set of all infinite sequences $(a_1, a_2, \dots)$ of integers $a_i$.  We can add elements of $G$ coordinate-wise, i.e.
\[(a_1, a_2, \dots) + (b_1, b_2, \dots) = (a_1 + b_1, a_2 + b_2, \dots).\]

Let $\mathbb{Z}$ denote the set of integers.  Suppose $f : G \rightarrow \mathbb{Z}$ is a function satisfying $f(x + y) = f(x) + f(y)$ for all $x$, $y \in G$.

Let $e_i$ be the element of $G$ with a 1 in position i and 0's elsewhere.  Show that $f(e_i) = 0$ for all but finitely many $i$.

I remember, we were solving a similar problem, where we knew that $f(e_i)=0$ for all $i$. The statement was to prove $f\equiv 0$. I think, it was from Iran.  :?

Just a trivial observation, which probably doesn't help anyway :). It suffices to show that $f(e_i)=0$ for at least one $i$.

[quote="grobber"]Just a trivial observation, which probably doesn't help anyway :). It suffices to show that $f(e_i)=0$ for at least ione $i$.[/quote]
 :rotfl:  :rotfl:  :rotfl:
I like your humor.

[Edit: Post irrelevant now that the previous post has been erased. ]

I know I am not so smart as you. I can't understand your last assertion for initial problem. I have some doubts.

[Edit. This 'solution' was totally erroneous. I've therefore deleted it. ]

Actually, here is a short solution:

Put $a_i = f(e_i)$, and assume to the contrary that $a_i \neq 0$ for $i \in I$, where $I \subset \mathbb{N}$ is an infinite set. Define the sequence $\{b_i\}$ of integers inductively by $b_1 = 0$, $b_i = 0$ if $i \notin I$, and $b_i = 2^{1 + \max\{|a_1b_1|,\ldots,|a_{i-1}b_{i-1}|\}} \mathrm{sgn} \, a_i$ if $i \in I$, where $\mathrm{sgn} \, x \in \{0, \pm 1\}$ is the sign of $x \in \mathbb{R}$.

Consider the $2$-adic number
\[
A := \sum_{i \geq 1} a_ib_i2^{i-1} \in \mathbb{Z}_2.
\]
By this construction, if $r_i = 1 + \max \{|a_1b_1,\ldots,|a_{i-1}b_{i-1}|\}$ and $i \in I$, then the $(r_i - 1)$st $2$-adic digit of $A$ is $0$ and the $r_i$th $2$-adic digit of $A$ is $1$. As $I$ is an infinite set, this shows in particular that the binary digits of $A$ do not stabilize (to neither $0$ nor $1$), so $A$ cannot be a rational integer. On the other hand, the additivity of $f$ implies that
\[
A = \sum_{i \geq 1}b_i2^{i-1}f(e_i) = f(b_1,2b_2,4b_3,\ldots) \in \mathbb{Z}
\]
is a rational integer, which is a contradiction.

--Vesselin

But vess! Why did you not write that your previous solution is wrong?
It is very unpleasant, in my opinion.

vess, are you allowed to sum $b_i2^{i-1}f(e_i)$ without knowing that all but a finite number of terms are $0$ in advance? I mean, you have a series whose terms are integers.. Not every element of $G$ can be written as an integral combination of $e_i$'s (in fact, somewhere on this forum there's a proof written by amfulger that $G$ is not a free $\mathbb Z$-module). Am I missing something here? :?

[Edit: I'm referring to vess' last message]

[quote="Myth"]But vess! Why did you not write that your previous solution is wrong?
[/quote]

That's why I typed this (new) solution. And it's definitely correct. Okay, I'll add the comment that the first solution is wrong. (Actually, the claim that if $\sum a_ip^i$ is a rational integer for all primes $p$ then $\{a_i\}$ stabilizes to zero is false, as I saw later).

Grobber, I could have written it in the following way: consider the $2$-adic integer
\[
A = \sum_{i \geq 1} a_ib_i2^{i-1} \in \mathbb{Z}_2,
\]
and let $A' = f(b_1,2b_2,4b_3,\ldots)$. For each $k \in \mathbb{N}$, we have that
\[
A' = \sum_{1 \leq i \leq k} a_ib_i2^{i-1} + 2^kf(0,0,\ldots,b_{k+1},2b_{k+2},\ldots) \equiv A \pmod{2^k}.
\]
By letting $k \to \infty$, we have $A' = A$.

I'm sorry, about that, I'll rephrase what I meant, since my previous question could be answered by saying that we're working with $p$-adics, so that series converges.

I don't think we are allowed to write $f(x_1,x_2,x_3,\ldots)=x_1f(e_1)+x_2f(e_2)+\ldots$ unless we actually know that all the $f(e_i)$ are null except for a finite number of terms (and this is what we want to prove :)). The additivity of the function can only be used for finite sums, I think. This has nothing to do with $p$-adic series; it's simply the statement that the element $(x_1,x_2,\ldots)$ is not generated by the $e_i$ unless only finitely many of the $x_i$ are non-null.

[Edit: I hadn't seen the response when I posted this :)]

O.K., can we move it to the 'Solved Problems' section now?

Btw, here are two interesting questions that I couldn't answer: what if we ask, in this and the Iranian problems, for rational-valued (instead of integer-valued) functions? That is,

[b]Problem 1. [/b] If $f: G \to \mathbb{Q}$ is additive and $f(e_i) = 0$ for all $i$, is it necessarily true that $f \equiv 0$ identically?

[b]Problem 2. [/b] If $f : G \to \mathbb{Q}$ is an additive function, is it necessarily true that $f(e_i) = 0$ for all but finitely many $i$?

--Vesselin

Hmm...
Consider a Hammel basis of $G$ over $\mathbb{Q}$, which include all $e_i$.
That's all.

Regarding the second one, couldn't we take $f(e_i)=\frac 1{2^{i-1}},\ \forall i\ge 1$, and define $f$ on all the other $0,1$ sequences as if it were countably additive?; take $f(x_1,x_2,\ldots)=x_1+\frac {x_2}2+\frac{x_3}{2^2}+\ldots$ for all sequences $(x_n)$ with elements $0$ or $1$. Since $G$ (as a $\mathbb Z$-module ) is generated by the sequenes $(x_n)$ with elements $0$ and $1$, we can complete the construction of $f$ to make it a homeomorphism.

Hope this isn't total nonsense...

Ooh, yes, of course.  :blush:

But don't you agree with the solution? Why not move it out of the unsloved section? :?  Or, what's wrong with my (new) solution?

[Edit: This post was addressed to Myth. ]

???
We never move solved (?) problems to Solved Section so soon. It will stay here for some time. Maybe someone will find a mistake... :roll:

[quote="Myth"] solved (?) problems [/quote]

And in what part of the proof (?) do you have any doubts? It seems perfectly obviously clear to me ... . If necessary, I can re-write it without using the notion of $p$-adic numbers ... Why's all that skepticism?

Vess!
I said you my opinion about this situation. It was very unpleasant to not mention that your solution was wrong. Though we knew about your mistake and we were waiting for your response for some days. And after that do you want we believe in your second solution? I am upset.

I know that your second solution doesn't require $p$-adic numbers.

[b]With regards[/b],
Myth

Okay, I'm sorry. I was abroad the past two days and couldn't post anything in this period, even though I knew the 'solution' is totally wrong. And that 'solution' was written at about 3 a.m. ...  And, no, I don't want you to [i]believe[/i] in the solution; I'm just saying it's obvious enough.

I know. But even I think it is correct it will stay here for some time.

It is hard for me to get an overview over this thread. Which proofs are correct, which not? What about a complete proof?

Hello Mr. Grothendieck :P (nice photo!) I just saw your post in this topic, and have deleted one of my early posts that may have confused you. My apologies. Everything else written in this thread is correct, and the proof in Post #9 is complete... well, almost -- correct it :)

Here is a similiar [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5090]post[/url] from Iran.


calculus
integration
limit
calculus computations

(1) For $ 0 < x < 1$, prove that $ (\sqrt {2} \minus{} 1)x \plus{} 1 < \sqrt {x \plus{} 1} < \sqrt {2}.$

(2) Find $ \lim_{a\rightarrow 1 \minus{} 0} \frac {\int_a^1 x\sqrt {1 \minus{} x^2}\ dx}{(1 \minus{} a)^{\frac 32}}$.

I would like have the solution without using L'Hopital. :wink:


Usually, you can get a formula for what integers satisfy such a condition, and if you can get the minimum and maximum integers, you can count them, or you can get a formula for which integers do not satisfy the conditions, then count the invalid integers in the range of all valid integers and then count that way. At least, that's what I've seen.

I think you can use the chicken mcnugget theorem. (search it up :) )

[quote="Truffles"]What do you usually do when you have a problem that's like, "How many distinct integers can be made by adding three integers of set S?" and can someone explain why the method works?[/quote] Can you give an example of a question and have us help you with it?


Find all integer solutions to the following equation:
                                   $x^{2}+7=2^{n}$

I gave a very ugly solution to this my first few weeks on this forum.  :D

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66245

I will let you know if I find a neater one.

Edit:  Oh, I forgot.  In fact, I incorrectly ruled out a case in that solution, so it is not complete.  :(


geometry
analytic geometry
perimeter
search
area of a triangle
number theory

write as many theorems as you can with explanation

geometry- I am beyond bad at theorems.

shoelace theorem- basically u list the coordinates x1 y1, then under it x2 y2, then under x3 y3, then x1 y1 under. Then add the multiplied diagonals facing one way and then subtract the multiplied diagonals facing the / way. then times 1/2 and get an answer.

it's for finding the area of a triangle knowing 3 points.

or pick's theorem can be used. number of dots in the circle and then + x/2 with x the number of points on the perimeter of the triangle, and -1.

We have about 23 threads that collect theorems and formulas. Possibly you want to take a look at the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Number Theory Formulary[/url] or search for the others.


floor function
search
calculus
geometric series
number theory unsolved
number theory

[quote="sbos"]1. Are there infinitely many numbers $n$ such that $\pi(n)\mid n$, where
$\pi(n)$ is the number of primes less than $n$?[/quote]
Try to search this forum. I remember I saw a beautiful proof for this fact.

Ok. Look it here http://www.mathlinks.ro/Forum/viewtopic.php?t=5132
It is really an excellent proof!

I'm not sure if it's the right statement, just because I don't know the proof. But I found that problem on the Internet and it's from American mathematical monthly.

[quote="Myth"]Indeed???? :huh:[/quote]
My idea was to show that besides "trivial" solutions (i.e. $\left\lfloor\pi^1\right\rfloor=3$, $\left\lfloor\pi^3\right\rfloor=31$ and $\left\lfloor\pi^4\right\rfloor=97$) there are no other ones, but that is not correct - $\left\lfloor\pi^{12}\right\rfloor=924269$, $\left\lfloor\pi^{73}\right\rfloor=1958577254745770740635072198655932631$, and the counter example from my last message.

What makes me think that the source and text are correct iss that I've seen this problem posted before, and the author of that topic cited the same source, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=impossible&t=1660]here[/url]. There's no solution, of course :).

[quote="sbos"]1. Are there infinitely many numbers $n$ such that $\pi(n)\mid n$, where
$\pi(n)$ is the number of primes less than $n$?

2. Prove that $\lfloor\pi^n\rfloor$ is a prime number for only finitely
many values of $n$.

I have no idea how to solve these problems.
For the first one natural answer is yes, I suppose,
but how to prove that? For the second problem one should
obviously know something about number $\pi$...[/quote]
It's well known that $\sum_{p \ prime} \frac 1 p$ diverges and $\frac 1 \pi<1$ then if exists infinite $n$ such that $[\pi^n]$ is prime then if we call this set of $n$, $S_\pi$, then
\[
\sum_{n\in S_{\pi}} \frac 1 {[\pi^n]}=+\infty,
\]
but $[\pi^n]\ge \pi^n$, and $\sum_{n\in \mathbb{N}}\frac{1}{[\pi^n]}$ converges, and $\pi^n>0$ and $S_\pi\subseteq \mathbb{N}$, then we get a contradiction!!!.

In fact the same argument is possible to use with $[e^n]$ and, in general,  every real number greater than 1.

Best regards.

RobertuX

[quote="sbos"]I think you are not right Robertux. How do you know that $\sum1/[\pi^n]=+\infty$ for  $n\in S_\pi$?[/quote]
Because any subsucession of a divergent (to $\infty$) sucession with extrictly positive coefficients always diverges (to $\infty$).
And if $n\in S_\pi$ then $[\pi^n]$ is prime.

[quote="Renshaw"][quote]but $[\pi^n]\ge \pi^n$, and $\sum_{n\in \mathbb{N}}\frac{1}{[\pi^n]}$ converges[/quote]

Do you mean $[\pi^n]\ge \pi^{n-1}$?

And what do you mean by "subsucession"?[/quote]
Well, you are right $[\pi^n]\ge \pi^{n-1}$ but with this change the proof is right too,  and a subsucession of $a_n$ is $a_{m(n)}$ which the property that $m(n+1)>m(n)$ and $m(n)$ is a positive integer, for instance $\{a_{2n}\}_{n\in N}$ and $\{a_{2^n}\}_{n\in N}$ are subsucessions of $\{a_n\}_{n\in N}$.

ok?

[quote="Renshaw"]ok.

Then what do you mean by "extrictly positive"?[/quote]
forget this one, only I try to avoid the oscillant limit for divergent series. In this problem I have applied positive coefficients series theory (see any Calculus book to read topics in this way).

Best regards and thx for your help to fix the proof.

RobertuX

Guys, you still did not understand? If you start a discussion with this man, you will find many incredible things! Yet, I would prefer not to see his remarks spreading on a nice forum such as number theory.

[quote="Renshaw"]Here's what I'm hung up on:

Let $a_n = \frac{1}{n}$.  Let $m(n) = n^2$.

$\sum_{n\in N}{a_n}$ diverges.

Do you mean to suggest that this implies $\sum_{n\in N}{a_{m(n)}}$ diverges?[/quote]
My result is not respect the $a_n$ is respect the index of the sum (in your case) I mean, if $a_n=\sum_{k=1}^{n}b_n$ then $a_{n^2}=\sum_{k=1}^{n^2}b_n$.

[b]Lemma:[/b] Any subsucesison of a sucession divergent (to infinity) with positive coefficients is divergent (to ininity)
[b]Proof:[/b] If $a_n \to +\infty$ then
\[
\forall M>0, \  \exists \, n_0 \ : \ n \ge n_0,  |a_n|>M,
\]
then for every inifity subset of $\mathbb{N}$, namely ${\{m(n)|}$ exists $n_1$ such that $m(n_1)>n_0$ then for every $n>n_1$ we have that $|a_{m(n)}|>M$ then diverges to infinity.
Q.E.D.

[b]Note: [/b] Of course $m(n+1)>m(n)$ for every $n$.

Best Regards  and sorry for u Harazy my proof is right and I feel sad for ur way to talk with the people.

RobertuX

[quote="Renshaw"]In that case I don't think we are dealing with a "subsuccession".

Suppose we let $b_n = $ the reciprocal of the $n$th prime number.  Next we define $a_n=\displaystyle \sum_{k=1}^n{b_k}$.
Now we let $d_n = $ the reciprocal of the $n$th prime that is of the form $[\pi^e]$, and we let $c_n =  \displaystyle \sum_{k=1}^n{d_k}$.
Now, it is true that $d_n$ is a subsuccession of $b_n$, since $d_n = b_{m(n)}$ where $m(n)$ is defined such that the $n$th member of the form $[\pi^e]$ is the $m(n)$th prime.

However, it is not true that $c_n$ is a subsuccession of $a_n$, because that would mean that $c_n = a_{m(n)}$ for some $m(n)$.  Since $c_1 = \frac 1 3 $ and $a_n$ never equals $\frac 1 3$, there can be no such $m(n)$.[/quote]
Please write clearly your comment, but the concept is the following, if you consider sucesisons, u can see that $a_n:=sum_{k=1}^n b_n$ is a sucession on $n$, then it's soo clear the proof of the lemma, no way remarks about that, maybe u use another expression for a subsucession. But TELL me  if you understand the fact which I proved in last quote. If this is OK, then I can re-write the quote of the proof for u.

Best Regards.

I'm fine with your lemma.  I just don't think you are applying it correctly.

You claim that $\displaystyle
\sum_{n\in S_{\pi}} \frac 1 {[\pi^n]}$ diverges because it is a "subsuccession" of $\displaystyle \sum_{p \ prime} \frac 1 p$.

In my last post I simply tried to show that this is not so.

[quote="Renshaw"]i. e.  $\displaystyle \displaystyle
\sum_{n\in S_{\pi}} \frac 1 {[\pi^n]} $ is [b]not[/b] a subsuccession of $\displaystyle \displaystyle \sum_{p \ prime} \frac 1 p $.[/quote]
Yes, it is. If $n\in S_\pi$ then, by definition, $[\pi^n]$ is a prime, then we can consider the orther in primes to consider the order in $S_\pi$, I mean, $S_\pi\subset \mathbb{N}$ and $N$ is well-ordered, then the elements $n\in S_\pi$ is a subsucession of the positive integers.

Firstly: $\displaystyle \sum_{n\in S_{\pi}} \frac 1 {[\pi^n]} = \frac 1 3 + \frac 1 {31} + \frac 1 {97} + ...$

and $\displaystyle \sum_{p \ prime} \frac 1 p = \frac 1 2 + \frac 1 3 + \frac 1 5 + \frac 1 7 + ...$.
Note that those are plus signs and not commas.  (Perhaps this will make clear my previous post).

Secondly:
[quote]It's well known that $\displaystyle \sum_{p \ prime} \frac 1 p$ diverges and $\frac 1 \pi < 1$ then if exists infinite $n$ such that $[\pi^n]$ is prime then if we call this set of $n$, $S_\pi$, then

$\displaystyle \sum_{n\in S_{\pi}} \frac 1 {[\pi^n]}=+\infty,$

but $[\pi^n] \geq \pi^n$, and $\displaystyle \sum_{n\in \mathbb{N}}\frac{1}{[\pi^n]}$ converges, and $\pi^n > 0 $ and $\displaystyle S_\pi\subseteq \mathbb{N}$, then we get a contradiction!!!.
[/quote]

What if I apply this same reasoning thus:
It's well known that $\displaystyle \sum_{k \ integer > 0} \frac 1 k$ diverges and $\frac 1 \pi < 1$ then if exists infinite $n$ such that $[\pi^n]$ an integer then if we call this set of $n$, $S_\pi$, then

$\displaystyle \sum_{n\in S_{\pi}} \frac 1 {[\pi^n]}=+\infty,$

but $[\pi^n] \geq \pi^n$, and $\displaystyle \sum_{n\in \mathbb{N}}\frac{1}{[\pi^n]}$ converges, and $\pi^n > 0 $ and $\displaystyle S_\pi\subseteq \mathbb{N}$, then we get a contradiction!!!.

Am I to conclude that $[\pi^n]$  is an integer for only a finite number of $n$?

[quote="Renshaw"]
Am I to conclude that $[\pi^n]$  is an integer for only a finite number of $n$?[/quote]
I feel happy, you understood the main idea. Yes, I tried to do something like that. The fact is use that one is a part of an geometric series and the other is that $\sum_{p\ prime} 1/p$ diverges.

[quote="Renshaw"]Do you understand?

I just used your line of reasoning to prove an absurdity.[/quote]
Well, don't care if you don't understand a very SIMPLE mathematical way to solve a problem, sorry for you and for all don't understand, but NO-one write a simple proof, I hope another write another way to solve this beautifull problem. Bye. I feel tired to prove freshman mathematical results.

RobertuX

RobertuX, could you please make your proof clearer cause I don't understand it, too (I understand your lemma, but not the way you are applying it).

So, we have the following successions:

$a_1=\frac 1 2$
$a_2=\frac 1 2+\frac 1 3$
$a_3=\frac 1 2+\frac 1 3+\frac 1 5$
$a_4=\frac 1 2+\frac 1 3+\frac 1 5+\frac 1 7$
...
and
$c_1=\frac 1 3$
$c_2=\frac 1 3+\frac 1 {31}$
$c_3=\frac 1 3+\frac 1 {31}+\frac 1 {97}$
...

Now it is clear that $a_n\rightarrow\infty$ and now, by applying your lemma, you conclude that $c_n\rightarrow\infty$, too. But $\{c_n\}$ is not a subsuccession of $\{a_n\}$ (is it?), so you can't apply your lemma, right?

Please don't be impatient, just help us to understand this.

Robertux, you said [quote]My result is not respect the a_n is respect the index of the sum (in your case) I mean, if a_n=\sum_{k=1}^{n}b_n then a_{n^2}=\sum_{k=1}^{n^2}b_n[/quote]
but why there is $b_n$ and not $b_k$ ?!

Please can you confirm that what you say is that:
If $\exists \Phi: \mathbb{N}\rightarrow \mathbb{N}, \Phi(n+1)>\Phi(n) \forall n\in\mathbb{N}$ and $\exists (a_n)_{n\in\mathbb{N}}$ such that $\sum_{k=0}^{+\infty} a_k=+\infty$ thus we have $\sum_{k=0}^{+\infty} a_{\Phi(k)}=+\infty$ ?

Thanks

Guys, this will never end. God, If only I had the power to make the useless posts vanish... (of course, this would reduce my posts to less than half :D ).

My, my ... .

Dear fleeting_guest, thank you for your timely remark in the 'Divisibility sequences' thread; I realize that what I did there is wrong, and I apologize for my arrogancy. You raised a few points that I agree with completely. But if you're reading these lines, you'll know what made me make my rash comment; I wanted to prevent some readers from ... (you know what).

This is a very intriguing problem, by the way! I'll see what I can do!  ;)  Besides, does anyone know whether a solution to this has appeared in the AMM?

Wow! Bojan, what primality test did you use to establish the primality of $\lfloor \pi^{317} \rfloor$?

Btw, can we even show that $\lfloor \pi^n \rfloor$ is composite for infinitely many $n$? I think this latter thing should hold with $\pi$ replaced by any number $> 1$, but I don't yet have an idea to show it. That is, I conjecture that there does not exist a $\lambda > 1$ for which $\lfloor \lambda^n \rfloor$ is prime for all $n$; this should be known in the context of the Mills theorem, but I couldn't find any references. What do you think, is this trivial? Maybe I miss something, but I don't find it at all trivial ...

[quote="vess"]
Wow! Bojan, what primality test did you use to establish the primality of $\lfloor \pi^{317} \rfloor$?[/quote]
I used Derive 5 program, and if I'm well informed it uses some mix of the prime number sieve, the Rabin-Miller primality test, and the Lucas primality test.

[quote="vess"]My, my ... .Dear fleeting_guest, thank you for your timely remark in the 'Divisibility sequences' thread; [...] [/quote]

Thanks, I was on my way to deleting it.  I am more interested in the other question from that thread --- you mentioned a known classification of such sequences; are there any examples on the list other than second-order Fibonacci-type sequences and elliptic divisibility sequences?  In particular, are there higher genus analogues of the elliptic sequences?

[quote] You raised a few points that I agree with completely. But if you're reading these lines, you'll know what made me make my rash comment; I wanted to prevent some readers from ... (you know what).  [/quote]

I understood that.  Any open forum will attract its share of "interesting" characters, psychiatric cases, enthusiastic 12-year-olds, jokers, etc.  It is easy enough to indicate specific mathematical errors rather than enter into a discussion of particular individuals, especially when we don't really know who is behind the username.

[quote]
This is a very intriguing problem, by the way! I'll see what I can do!  ;)  Besides, does anyone know whether a solution to this has appeared in the AMM?  [/quote]

I guess this is true for any number satisfying a Diophantine condition of boundedly-good approximation by rationals, but don't have anything like a proof at the moment.

I have found a really interesting fact which in my opinion has a connection with problem 2). Namely, in "Introduction to theory of numbers" (2004) by Leo Moser, page 84, the following two problems are listed as [u]open[/u] (and attributed to Erdos):

Problem 23: Does $\left\lfloor e^n\right\rfloor$ represent infinitely many primes?

Problem 24: Does $\left\lfloor e^n\right\rfloor$ represent infinitely many composite numbers?

I believe that the same approach should be used for these two problems as for the our problem. Maybe it is also open?

[quote="sbos"]2. Prove that $\lfloor\pi^n\rfloor$ is a prime number for only finitely
many values of $n$.[/quote]
This problem appeared as problem 6212 in AMM 85 (1978), p389, and has not yet been solved.


hi all of the friends...

i have seen som of the the geometry problems from poland which are very nice...

could any body please help me and write some of the hardest geometry problems (poland) for me?

thanks alot dears ;) ...

[quote="neworder"]Do you mean problems from this year Polish Olympiad?[/quote]

not only this year...

but not too far ...

recent years.... ;)

[quote="zibi"]It depends on how we will define word "nice" :P

This one is from present Math olympiad:

http://www.mathlinks.ro/Forum/viewtopic.php?p=386912&highlight=#386912[/quote]

thanks alot $zibi$ but i have already solved it :blush: .

do u know the problems of ure contests olypiads...for example (the final round of poland olympiad 2005)

i really want to know how hard they are... ;)  ;)

thanks again...

[quote="TomciO"]http://www.mathlinks.ro/Forum/resources.php?c=137&cid=39&year=2005

Geometry from the second day has been solved by only one person IIRC.[/quote]

a very very very very nice problem...

this is one of the best problems which i saw...

but  :blush:  :(  :oops:

again i should say i had solved it...by inversion.

really inversion is incridible isnt it? ;)

i can send the solution, if u want it?

(summery:inversion with center $S$ and radius $SA.SD$)

oh i made a mistake it seems that the problem which i talked about it,is from day one... :P

ok i will think a bout it...

thanks dear tomico ;) ...

Hehe, no problem :).

[quote]i can send the solution, if u want it? [/quote]
Sure, post it here [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=337819#p337819[/url] because there isn't solution with the inversion.

[url]http://www.mathlinks.ro/Forum/resources.php?c=137&cid=39[/url]

Here you have many problems from finals of the our MO, if you were interested in seeing solutions to some problems or sth I can translate it for you...

[quote="TomciO"]

[url]http://www.mathlinks.ro/Forum/resources.php?c=137&cid=39[/url][quote]

thanks it was really use full....

about six or seven houres later i will post the solutions... :D

dears (tomice ,zibi, ....)

i will be glad if u write ure biographies...

are u classmate...?

thanks again... :lol:

from the time which i create this topic u really helped me.and i want to be friend with u...can i?

dear zibi u can see the solution in: http://www.mathlinks.ro/Forum/viewtopic.php?p=406041#406041

Thank you for posting this solution. I find solution to this problem using standard methods(some angle chase), but this using inversion is better  :)


[color=blue]Thấy trong box giới thiệu 1 cuốn sách b đ t, đây là một bài về 1 cuốn khác, bài này copy từ ddth. [/color] :D
[quote]
Như các bạn đã biết, cuốn sách Sáng Tạo Bất đẳng thức bản tiếng anh (Secrets in Inequalities) đã được chính thức xuất bản ở GIL Publishing house. Trong khoảng thời gian Sáng Tạo Bất đẳng thức được xuất bản đến nay, hungkhtn đã off trên diễn đàn toán học, cũng ít tham ra Mathlinks, thế nên có thể nói cuốn Secrets In Inequalities ra đời... thầm lặng và cũng không có nhiều bạn gửi bài đóng góp cho nó nếu so sánh với bản tiếng Việt (chỉ có 1 số người bạn thân và hungkhtn liên hệ trực tiếp qua mail).

Tuy cuốn Secrets in Inequalities I đã được chính thức xuất bản, Volume II đã hoàn thành xong, nhưng hiện tại cũng còn một khoảng thời gian trên dưới 1 tháng để sửa chữa và bổ sung. Vì vậy hungkhtn hi vọng trong khoảng thời gian này, các bạn sẽ tích cực đóng góp thêm những bài toán mới, chuyên đề mới, để cuốn sách thực sự là sản phẩm của học sinh Việt Nam.

Trong Volume 2 có một đặc điểm rất đặc biệt (mà hiếm thấy một cuốn sách khác tương tự), đó là sẽ có 1 chương riêng mang tên "The readers' Contributions", dành cho các đóng góp của tất cả mọi người trong và ngoài nước. Hiện tại chương này (Chương 7) đã có 2 bài viết rất thú vị, mỗi bài khoảng 25 trang và hi vọng trong thời gian tới nó sẽ dày thêm. Nếu bạn có bài viết trong chương này bạn sẽ được tặng một cuốn sách kỷ niệm.

Ngoài ra, sách sẽ được xuất bản ở Gil Publishing House (một nhà xuất bản rất nổi tiếng về Toán Olympiad). Có thể GIL còn hợp tác thêm với 1 nhà xuất bản... rất rất nổi tiếng khác để cùng xuất bản cuốn sách, vì thế có thể nói, một bài viết chuyên đề của bạn trong chương 7 còn là một vinh dự cho các bạn.

Hiện tại Hungkhtn cũng có một số project khác hợp tác với GIL về bất đẳng thức (chẳng hạn viết một số chuyên đề gửi báo Gazetta mathematica, Reflection...). Các bạn có thể liên hệ với mình để contribute cho các project này.

Cuối cùng, hungkhtn xin đưa lên bìa sách của cuốn Secrets in Inequalities. Phần Preface sẽ được đưa lên trong thời gian sớm nhất. Các thông tin khác các bạn có thể tìm thấy trong Diễn Đàn bất đẳng thức Việt Nam VIF.
____________________________

PS. Nếu bạn muốn propose bài toán cho cuốn sách tập 2, bạn có thể gửi trực tiếp đề bài và lời giải lên topic này. Hungkhtn mong nhận được sự cộng tác của mọi người.
[/quote]

Hôm trước thấy Hùng bảo là có Vol 1 rồi, hình thức thì đẹp hơn, không biết nội dung có gì mới không chú?

[quote="pvthuan"]Vo Danh Khach kia huyền bí quá. Một lần nhắn tin hỏi danh tính rồi mà không thấy hồi âm.[/quote]
Ngay cạnh nhà tôi, vừa nãy lão ấy dùng máy của tôi để post bài đấy chứ.  :lol:


function
trigonometry
algebra unsolved
algebra

Instead of 2005, use the more general n.

Put $a=\pi/2^n$.  Then note that $-\sin(3x) / \sin(x) = 4 \sin(x)^2 -3$.  Hence the expression, without the first term, then becomes $\sin(3a) \sin(6a) \sin(9a) \cdots / \sin(a) \sin(2a) \sin(3a) \cdots$.

Then simply convert all of the sines in the numerator to sines in the interval $[0,\pi/2]$ using the identities $\sin(x)=\sin(\pi-x)$ and $\sin(x)=-\sin(\pi+x)$ where appropriate.  No two reduced values are equal (since otherwise their sum or difference would be a multiple of $\pi$, impossible), and there are $2^{2004}+1$ of them, so they are ${a, 2a, 3a, \cdots, 2^n a}$ in some order.  Hence the product is 1 or -1, depending on how many of the sines in the numerator are negative.  A little thought shows that when n is even, there are an odd number of negative sines, and when n (>1) is odd there are an even number of negative sines.  Hence, adding back in the first term of -3, the answer for a general n is $3 (-1)^n$.  When n is 2005, the answer is -3.


Paper and the Moon

A sheet of paper is .016 cm thick. Suppose that you tear this paper in half. Then you stack the two halves together and tear them in half. Then you take the four pieces, stack them, and rip them in half. If it were possible, you would continue this process of stacking the ripped pieces together and tearing them apart 60 times.
You need to answer the following four questions correctly to receive credit. All answers must be written in scientific notation.

1.How many pieces would the stack have?

2.How high would the stack reach?

3.Would it be taller than the highest building (not under construction)? How high is the tallest building?

4.Would it be higher than the moon at its greatest distance from Earth? What is this distance?

Be sure to show all your work and explain how you got your answers.

Give the height of the tallest building (or don't ask the question in the first place). We are lazy. (We also aren't architects/researchers.)

1) 1.153E+18 pieces

No way Alex! This doesn't look like a homework problem.  :rotfl:

The tallest building is 1,670 ft.

The farthest the moon can go away from the Earth is 252,710 mi.

[quote="probability1.01"]*cough* homeworkhelp *cough*[/quote]

no not exactly...I did get this in my Math class but it was just for fun...she just wanted to see if we were smart enough to figure it out...but I do have the answers to it...

Now here's MY question...
How did this get into Pre-Olympiad?

1) We just plug $2^{60}$ into a calculator which spits out our answer.
2) We just multiply the previous answer by the thickness, $.016$cm, or probably more appropriately $1.6*10^{-4}$m (assuming ideal conditions) using a calculator...or we just put into our calculators $2^{64}*10^{-5}$m and round to sig figs.
3&4) We just compare these values to those in 2.

It isn't something which takes a whole lot of thinking, but rather calculator work...

[quote="Altheman"][quote="K81o7"]It isn't something which takes a whole lot of thinking, but rather calculator work...[/quote]

welcome to my math class :P[/quote]

Your math class is like that?

...

[quote="K81o7"]It isn't something which takes a whole lot of thinking, but rather calculator work...[/quote]

Don't you just love abstract algebra?

Whatever it is, it is not pre-olympiad stuff. Feel free to re-post it in the correct forum. [I guess they'll be enjoyed with that in beginner section, but am not permitted to move it to there]


geometry
geometric transformation
rotation
function
combinatorics unsolved
combinatorics

$d$ and $n$ are positive integers such that $d \mid n$. The n-number sets $(x_1, x_2, \cdots x_n)$ satisfy the following condition:
(1) $0 \leq x_1 \leq x_2 \leq \cdots \leq x_n \leq n$

(2) $d \mid (x_1+x_2+ \cdots x_n)$

Prove that in all the n-number sets that meet the conditions, there are exactly half satisfy $x_n=n$.

To reformulate we need to show that the number of partitions of numbers divisible by $d$ into $n$ parts from $0$ to $n-1$ equals the number of partitions of numbers divisible by $d$ into $n-1$ parts from $0$ to $n$ (this is done after we remove one $n$). We can actually construct a bijection which is well-known Ferrer Diagrams: represent the partition graphically on the lattice : firstly put the largest part, then next put the second largest part and so on. The bijection is the rotation by 90 deg. Or, if you want it more algebrically,if we have $p_n$ parts equal to $n$, $p_{n-1}$ parts equal ro $n-1$ and so on, we associate the sysetm $p_n, p_n+p_{n-1}, p_{n}+p_{n-1}+p_{n-1}, \ldots$

Another solution based on generating function :)

This also appeared on https://artofproblemsolving.com/community/c6h1520134p9154743 .


I remember somewhere, we'd discussed the possibility of doing an ARML seminar not unlike the one for AIME. What's the status quo on that?

June 3rd and 4th.  http://www.arml.com

As for ARML, I might start a topic like the AMC a day's and AIME a day's I did where I post a problem every day or so.  I'm a bit busy now though...

I haven't heard anything about the seminar idea but I highly doubt they have gotten very far on one if they did start making it.  They seem to have been really busy lately.  But you'll have to wait for an instructor for a definitive answer.

I believe the Texas team is leaving on Wednesday, June 1, and coming back Sunday, June 5.

joml88, I could run your thread until you free up since I (as I imagine you would too) have some old ARML problem books that I haven't even touched yet.

I only have the 1995-2003 one that they sold at last years ARML.  If you have some of the older ones that would be great.  We could take turns each week or so to make it easier on both of us.

Actually, I think I'll probably post the first problem tonight in the Intermediate Forum.  I would do the Other US Contests one but I think more people visit the Intermediate.

Yes, I also have a copy of the 1989-1994 ARML-NYSML contests.

Right...I was thinking the same. Sounds good!  :)

I'd stop that until you have permission from the proper people to post ARML problems. There's a reason you had to buy the book to get the problems.

I'm not 100 percent certain, but I believe fair use allows you to replicate any one math problem for educational purposes.  It might not be okay however to reproduce full tests on the forums.

As for the ARML Problem Series, that's something we have discussed and may come to fruition sometime in the future, but not this year.

Remember that ARML teams can use the local forums to create practice sessions among their own teams.  Part of the inspiration for the local and state forums was to help ARML teams establish active online discussions.

In my case, I tend to use this material (and yes, I have both the '89-'94 and '95-'03 books that joml88 and towersfreak2006 referred to above) primarily for "live" or "real-time" practices. We do have practices in which 70% or more of the team(s) meets in a central location once or twice a month. I would suspect that's not true for the Texas team, with Dallas being >200 miles from Houston.

In live practice, we get to try out our teamwork and our group strategies, things that are difficult to emulate on line.

For those reasons, although I would be OK with occasional scattered problems, I would prefer not to see whole team rounds, power rounds, or relays posted.

If students are looking to work ARML problems without depleting the supply of potential practice tests, they might do well to check out the problems from the HMMT which are posted online.  While the topics covered are not exactly the same, many of them are and the difficulty seems appropriate.


floor function
function
modular arithmetic
algebra

Is that; what is the $ 163^{rd}$ term?
[hide]
Suppose the $ 163^{rd}$ term is $ n$, then if we take away all the multples of $ 2$ and $ 3$ from $ n$ then we should be left with $ 163$. That is;

$ n \minus{} \left\lfloor \frac {n}{2} \right\rfloor \minus{}\left \lfloor \frac {n}{3} \right\rfloor \plus{} \left\lfloor \frac {n}{6} \right\rfloor \equal{} 163$

First it is easiest to check without the floor function
$ n \equal{} 489$

Plugging it back in we see
$ 489 \minus{} \lfloor 244.5 \rfloor \minus{} \lfloor 163 \rfloor \plus{} \lfloor 81.5 \rfloor \equal{} 163$, so that's our number[/hide]

[quote="ocha"]Is that; what is the $ 163^{rd}$ term?
[hide]
Suppose the $ 163^{rd}$ term is $ n$, then if we take away all the multples of $ 2$ and $ 3$ from $ n$ then we should be left with $ 163$. That is;

$ n \minus{} \left\lfloor \frac {n}{2} \right\rfloor \minus{} \left \lfloor \frac {n}{3} \right\rfloor \plus{} \left\lfloor \frac {n}{6} \right\rfloor \equal{} 163$

First it is easiest to check without the floor function
$ n \equal{} 489$

Plugging it back in we see
$ 489 \minus{} \lfloor 244.5 \rfloor \minus{} \lfloor 163 \rfloor \plus{} \lfloor 81.5 \rfloor \equal{} 163$, so that's our number[/hide][/quote]

Sorry,
$ 489$ is divisible by $ 3$

[quote="PKMathew"]Delete all multiples of $ 2$ and multiples of $ 3$ from $ 1,2,3,4,5,6\cdots$ natural numbers , what is the number of 163 rd in the rest  list?[/quote]

Of the first 300 natural numbers, only $ 300\minus{}150\minus{}100\plus{}50\equal{}100$ are on the list.

Of the next 100, there are $ 100\minus{}50\minus{}33\plus{}16\equal{}33$ numbers.

Of the next 100, there are $ 33$ numbers.

Thus, the 166th number is 499. The 165th is 497, 164th is 493, and 163rd is 491.

[quote="BOGTRO"][quote="PKMathew"]Delete all multiples of $ 2$ and multiples of $ 3$ from $ 1,2,3,4,5,6\cdots$ natural numbers , what is the number of 163 rd in the rest  list?[/quote]

Of the first 300 natural numbers, only $ 300 \minus{} 150 \minus{} 100 \plus{} 50 \equal{} 100$ are on the list.

Of the next 100, there are $ 100 \minus{} 50 \minus{} 33 \plus{} 16 \equal{} 33$ numbers.

Of the next 100, there are $ 33$ numbers.

Thus, the 166th number is 499. The 165th is 497, 164th is 493, and 163rd is 491.[/quote]
[hide]
For any  natural number $ n\equiv \{0,1,2,3,4,5\}\pmod6$

Only $ \{1,5\}$ may be matched with the question.

When $ n\equiv 1\pmod6$,

Then $ n \equal{} 1,7,13,19,25......$(sq 1)

When $ n\equiv 5\pmod6$,

Then $ n \equal{} 5,11,17,23,29......$(sq 2)

Since $ \frac {163}{2} \equal{} 81.5$

Consider each sequence reaches $ 81^{st}$~$ 82^{nd}$ term.

$ 1 \plus{} (81 \minus{} 1)\times 6 \equal{} 481$ ; $ 5 \plus{} (81 \minus{} 1)\times 6 \equal{} 485$

 $ 1 \plus{} (82 \minus{} 1)\times 6 \equal{} 487$ ; $ 5 \plus{} (81 \minus{} 1)\times 6 \equal{} 491$

When two seqences merge with 81 terms each other up to 162 terms,
then the $ 162^{nd}$ term $ \equal{} 485$ , the $ 163^{rd}$ term $ \boxed{ \equal{} 487}$ [/hide]

[hide]If we write down the sequence $ (1, 5, 7, 11, 13...)$, we can see that $ a_{2k}\equal{}a_{2k\minus{}1}\plus{}4$ and $ a_{2k\plus{}1}\equal{}a_{2k}\plus{}2 \implies a_{2k\plus{}1}\equal{}a_{2k\minus{}1}\plus{}6 \implies a_{2k\plus{}1}\equal{}a_1\plus{}6k \implies \boxed{a_{163}\equal{}487}$[/hide]


function
inequalities
trigonometry
logarithms
inequalities proposed

Let $a,b$ be positive reals such that $\left(0\leqq a<\frac{\pi}{4},0\leqq b<\frac{\pi}{4}\right)$. Prove the following inequality.

\[\sqrt{\tan a\cdot\tan b}\leqq \tan \left(\frac{a+b}{2}\right)\leqq \frac{1}{2}(\tan a+\tan b)\]

It is just Jensen for $f(x)=\ln \tan x$ (first inequality) and for $g(x)=\tan x$ (second one).

That's right.

Can anyone have another solution?

kunny

(i)1-tg((a+b)/2)^2=cos(a+b)/cos((a+b)/2)^2
1-tg(a)tg(b)=cos(a+b)/(cosacosb)
and cos(a+b)>0
2cosa.cosb=cos(a-b)+cos(a+b)<=1+cos(a+b)=2cos^2((a+b)/2)

(ii)tg(a)+tg(b)=sin(a+b)/(cosa.cosb)
2tg((a+b)/2)=sin(a+b)/cos((a+b)/2)^2
and cos((a+b)/2)^2>=cosa.cosb

We have result!

The actual purpose of this post is to give a lemma to a proof of a triangle inequality.

In fact, let's generalize Kunny's assertion:

[color=blue][b]Theorem 1.[/b] Let x, y, x', y' be four numbers in the interval ]-90°; 90°[ such that $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}$ and $\left|x-y\right|\geq\left|x^{\prime}-y^{\prime}\right|$. Then,

$\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$.

Additionally, if $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}\leq 90^{\circ}$, then

and

$\tan x\tan y\leq\tan x^{\prime}\tan y^{\prime}$.[/color]

[i]Note.[/i] This quickly yields the assertion of Kunny's problem if we set x = a, y = b and $x^{\prime}=y^{\prime}=\frac{a+b}{2}$ (in fact, $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}\leq 90^{\circ}$ is obvious; it is equally obvious that $\left|x-y\right|\geq\left|x^{\prime}-y^{\prime}\right|$, as x' = y' yields $\left|x^{\prime}-y^{\prime}\right|=0$; now, $\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$ and $\tan x\tan y\leq\tan x^{\prime}\tan y^{\prime}$ become $\tan a+\tan b\geq 2\tan\frac{a+b}{2}$ and $\tan a\tan b\leq\tan^2\frac{a+b}{2}$).

[i]Proof of Theorem 1.[/i] First, we will prove that $\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$ if $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}$.

Denote x + y = x' + y' = d. Then, the condition $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}$ becomes $0^{\circ}\leq d$, while on the other hand, we have d = x + y < 180° (since both numbers x and y are from the interval ]-90°; 90°[). Hence, sin d is nonnegative. Also, since the numbers x and y are from the interval ]-90°; 90°[, their cosines cos x and cos y are positive; thus, the product cos x cos y is also positive. But

$\cos x\cos y=\frac12\left(\left(\cos x\cos y-\sin x\sin y\right)+\left(\cos x\cos y+\sin x\sin y\right)\right)$
$=\frac12\left(\cos\left(x+y\right)+\cos\left(x-y\right)\right)=\frac12\left(\cos d+\cos\left(x-y\right)\right)$.

Hence, cos d + cos (x - y) is positive. Now,

$\tan x+\tan y=\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}=\frac{\sin x\cos y+\sin y\cos x}{\cos x\cos y}=\frac{\sin\left(x+y\right)}{\cos x\cos y}$
$=\frac{\sin d}{\frac12\left(\cos d+\cos\left(x-y\right)\right)}=\frac{2\underbrace{\sin d}_{\geq 0}}{\underbrace{\cos d+\cos\left(x-y\right)}_{>0}}$.

Similarly,

$\tan x^{\prime}+\tan y^{\prime}=\frac{2\underbrace{\sin d}_{\geq 0}}{\underbrace{\cos d+\cos\left(x^{\prime}-y^{\prime}\right)}_{>0}}$.

Hence, in order to prove that $\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$, it is enough to show that $\cos\left(x-y\right)\leq\cos\left(x^{\prime}-y^{\prime}\right)$. This is obviously equivalent to $\cos\left|x-y\right|\leq\cos\left|x^{\prime}-y^{\prime}\right|$, what is true since the numbers $\left|x-y\right|$ and $\left|x^{\prime}-y^{\prime}\right|$ lie in the interval [0°; 180°] (since the numbers x, y, x', y' lie in the interval ]-90°; 90°[) and satisfy $\left|x-y\right|\geq\left|x^{\prime}-y^{\prime}\right|$, while the cosine function is monotonically decreasing in this interval. Thus, the inequality $\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$ is proven.

Now, we will show that $\tan x\tan y\leq\tan x^{\prime}\tan y^{\prime}$ under the assumption $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}\leq 90^{\circ}$. In fact, since d = x + y, this assumption yields $0^{\circ}\leq d\leq 90^{\circ}$, so that tan d is nonnegative.

by the tangens addition formula, $\tan d=\tan\left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$, so that $1-\tan x\tan y=\frac{\tan x+\tan y}{\tan d}$. Similarly, $1-\tan x^{\prime}\tan y^{\prime}=\frac{\tan x^{\prime}+\tan y^{\prime}}{\tan d}$. Since tan d is nonnegative, we thus can use the above proved inequality $\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$ to obtain $1-\tan x\tan y\geq 1-\tan x^{\prime}\tan y^{\prime}$, so that $\tan x\tan y\leq\tan x^{\prime}\tan y^{\prime}$.

Thus, Theorem 1 is completely proven.

  Darij

[quote="hungkhtn"]$\mathbf{(i)}\ 1-\tan^{2}\left(\frac{a+b}{2}\right)=\frac{\cos\left(a+b\right)}{\cos^{2}\left(\frac{a+b}{2}\right)}$
$1-\tan(a)\tan(b)=\frac{\cos(a+b)}{\cos a\cos b}$
and $\cos(a+b)>0$
$2\cos a\cos b=\cos(a-b)+\cos(a+b)\leq 1+\cos(a+b)=2\cos^{2}\left(\frac{a+b}{2}\right)$

$\mathbf{(ii)}\ \tan(a)+\tan(b)=\frac{\sin(a+b)}{\cos a\cos b}$
$2\tan\left(\frac{a+b}{2}\right)=\frac{\sin(a+b)}{\cos^{2}\left(\frac{a+b}{2}\right)}$
and $\cos^{2}\left(\frac{a+b}{2}\right)\geq \cos a\cos b$

We have result![/quote]


inequalities
geometry
3D geometry
induction
calculus
integration

For any positive integer n,prove that $ (1\plus{} \frac {1}{n})^n < (1\plus{}\frac{1}{n\plus{}1})^{n\plus{}1}$.
Thanks.

hello, using the inequality $ \sqrt[n\plus{}1]{ab^n}<\frac{a\plus{}nb}{n\plus{}1}$ we get with $ a\equal{}1,b\equal{}1\plus{}\frac{1}{n}$
$ \sqrt[n\plus{}1]{1\cdot\left(1\plus{}\frac{1}{n}\right)^n}<\frac{1\plus{}n(1\plus{}\frac{1}{n})}{n\plus{}1}\equal{}\frac{n\plus{}2}{n\plus{}1}\equal{}1\plus{}\frac{1}{n\plus{}1}$ therefor we get $ \left(1\plus{}\frac{1}{n}\right)^n<\left(1\plus{}\frac{1}{n\plus{}1}\right)^{n\plus{}1}$.
Sonnhard.

[quote="nobelium"]For any positive integer n,prove that $ (1 \plus{} \frac {1}{n})^n < (1 \plus{} \frac {1}{n \plus{} 1})^{n \plus{} 1}$.
Thanks.[/quote]

Just interested, that expression somehow reminds me of $ e$; can this variable possibly be used to determine a solution to this inequality?

[quote="Dr Sonnhard Graubner"]using the inequality $ \sqrt [n + 1]{ab^n} < \frac {a + nb}{n + 1}$
[/quote]

This approach doesn't belong in High School Basics.  You stated something
known to you, but I'm sure it is virtually unknown to others at this level.

If you had given a derivation as to how that formula comes about, then I
could see that longer post fitting.
------------------------------------------------------------------------------------------------
EDIT:

$ (1 + \frac1n)^{n} < (1 + \frac{1}{n + 1})^{n + 1}$

Divide both sides by the left-hand side expression:

$ \frac{(1 + \frac1n)^n}{(1 + \frac1n)^n} < \frac{(1 + \frac{1}{n + 1})^{n + 1}}{(1 + \frac1n)^n}$

$ 1 < \frac{(1 + \frac{1}{n + 1})^n}{(1 + \frac1n)^n}\cdot{(1 + \frac{1}{n + 1})}$

$ 1 < (\frac{n^2 + 2n}{n^2 + 2n + 1})^n\cdot{(1 + \frac{1}{n + 1})}$

$ 1 < [1 - \frac{1}{(n + 1)^2}]^n\cdot(1 + \frac{1}{n + 1})$ . . . (*)

Using the form $ (a + b)^n$, $ [1 - \frac{1}{(n + 1)^2})]^n \ge (1)^n + n(1)^{n - 1}[\frac{1}{(n + 1)^2}]^1$

$ [1 - \frac{1}{(n + 1)^2)}]^n \ge  1 - \frac{n}{(n + 1)^2}$

Substituting this into (*) above:

$ 1 < [1 - \frac{n}{(n + 1)^2}](1 + \frac{1}{n + 1})$

$ 1 < 1 + \frac{1}{n + 1} - \frac{n}{(n + 1)^2} - \frac{n}{(n + 1)^3}$

Combine the three fractions into one fraction:

$ 1 < 1 + \frac{1}{(n + 1)^2}$

For the proof, start with...

$ 1 < \frac{1}{(n + 1)^2}$ . . . . and reverse the steps.

err.. here's my attempt at proving it intuitively:
By adding a 1 to the fraction, we kind of balance out the the expression inside the parentheses. So, no matter how big n+1 gets, the 1 would always be there to balance it out. But here we should notice, that 1/n > 1/n+1 (obvious since the denominator is getting larger). But now, we have the exponent. This exponent does ALOT more in changing (in this case increasing) the value than does the +1. Now, squares, cubes, etc. have a strange pattern of increments.
For example, if we take n=1, we arrive at 2<9/4 which is true. The reason being, that the 1 balances out the inner expression, and since for the 2nd part (in this case we have 3/2) has a funny pattern, and for any numbers that are that close to each other, like 2 and 3/2, just a simple difference in the exponent will really drastically change the numbers. Sure for my example, 2 and 9/4 are close in value, and in fact, if we let n=2, then the two expressions would be even closer. But its really important to realize that when we balance out the fraction, the numerator > denominator, and to whatever degree it is raised to, the numerator becomes even more greater than the denominator. And if that degree is changed by adding 1 more to, then that numerator will consequently become even more larger. All in all, the balancing nature of 1 and the increase in the raised degree have big effects on the values of the expressions.

Sorry, I really cant think of a mathematical (numerical representation, induction etc. ) to prove this, but intuition is just as important.

[quote="nobelium"]Can anyone please prove this:
Given a<b, $ (n \plus{} 1)a^n < b^n \plus{} ab^{n \minus{} 1} \plus{} ... \plus{} a^n < (n \plus{} 1)b^n$

Thanks again.[/quote]

since $ a<b$ we have $ a^n<ab^{n\minus{}1}<b^n$
$ a^n<a^2b^{n\minus{}2}<b^n$
$ \cdots$
$ a^n<a^{n\minus{}2}b^2<b^n$
$ a^n<a^{n\minus{}1}b<b^n$
easily we get $ (n \plus{} 1)a^n < b^n \plus{} ab^{n \minus{} 1} \plus{} ... \plus{} a^n < (n \plus{} 1)b^n$

Thanks a lot,friend.

And I agree with you.Just like Jammu and Kashmir and Aksai Chin are parts of India,Taiwan too remains an integral part of China.


logarithms
inequalities unsolved
inequalities

Substitute $ a \equal{} x \plus{} y$ and so on. Then you get
\[ \sum \frac {x \plus{} y}{x \plus{} 3y} \ge \frac {3}{2}\]
or
\[ \frac {3}{4} \ge \sum \frac {1}{e^a \plus{} 3},\]
where $ \frac {y}{z} \equal{} e^a$ and so on. Hence $ a \plus{} b \plus{} c \equal{} 0$. Then you have to distinguish 3 cases ($ a,b,c \le \ln 3$, $ a,b \le \ln 3$, $ a \le \ln 3$, where $ a \le b \le c$). Every case can be done by Karamata, noting that $ f(t) \equal{} \frac {1}{3 \plus{} e^t}$ is concave on $ [ \minus{} \infty, \ln 3]$ and convex on $ [\ln 3, \infty]$ and Am-Gm in the end.

I used $ a, b, c$ twice, but they are not the same…^^ I just recognised that after writing the post...

we can setting $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x,x,y,z > 0$, that we have\[ \sum {\frac {a}{{2a \plus{} b \minus{} c}}} \equal{} \sum {\frac {{x \plus{} y}}{{x \plus{} 3y}}} \equal{} 3 \minus{} 2\sum {\frac {y}{{x \plus{} 3y}}}\]so\[ \sum {\frac {a}{{2a \plus{} b \minus{} c}}} \ge \frac {3}{2}\]\[ \Leftrightarrow 3 \minus{} 2\sum {\frac {y}{{x \plus{} 3y}}} \ge \frac {3}{2}\]\[ \Leftrightarrow \sum {\frac {1}{{\frac {x}{y} \plus{} 3}}} \le \frac {3}{4}\]let $ \frac {x}{y} \equal{} t,\frac {y}{z} \equal{} u,\frac {z}{x} \equal{} v$, that we have$ t,u,v > 0$ and $ tuv \equal{} 1$, just need to prove that\[ \sum {\frac {1}{{t \plus{} 3}}} \le \frac {3}{4}\]\[ \Leftrightarrow 3\left( {t \plus{} u \plus{} v} \right) \plus{} 5\left( {tu \plus{} uv \plus{} vt} \right) \minus{} 24 \ge 0\]which is ture by Am-Gm.

It's posted in here: [url]http://www.mathlinks.ro/viewtopic.php?p=1559251#1559251[/url]


Does someone know how to prove a language is context-free using a pumping lemma? What would be the proof that the following languages are context-free?

1. Prove that

L = {x ∈ {a, b}* | count(b, x) = (count(a, x))^2}

is not a CFL.

2. Prove that

L = {(a^i)(b^j)(c^k) | i < j < k}

is not a CFL.

You want to prove that the language is [b]not[/b] context-free, right? OK, I will suppose that you know what pumping lemma for CF languages says. For every CF language, there exist constants p, q such that... p is the lower bound for word length and q is the upper bound for the substring. I will also use r which I define as the greater of p, q.

1. Suppose you have a word with r 'a's and r^2 'b's. You can find a substring xuy such that its length is < q <= r and that changing it to xxuyy will give a word from the same language. However, notice that there is at least one 'a' or 'b' in xy so the count of 'a's and 'b's will increase - 'a's at least by 1 and 'b's at least by (r+1)^2 - r^2 = 2r +1. However, the length of xy is less than r so the count of 'b's can't increase by that many. Therefore L is not CFL.

2. Let's take the word $a^rb^{r+1}c^{r+2}$ which is from L. There exists some xuy shorter that r, therefore xy can't contain both 'a's and 'c's. If it contains only 'a's then substituting xuy for xxuyy creates a word with count(a)>=count(b). If it contains only 'b's the situation is the same. If it contains only 'c's then substituting xuy for u gives a word which has count(b)>=count(c). If xuy contains both 'a's and 'b's then xxuyy will cause count(b)>=count(c). If xuy contains 'b's and 'c's then xuy->u will make the word have count(b)<=count(a). L is not CFL.

Sorry to intrude, but can anyone help me understand the proof of pumping lemma for cfl's. For regular languages it was too easy but I don't feel comfortable with the same in case of cfl's.

Thank you for your help...

The proof I know works something like this: First, take a CF grammar $G$ and remove all the epsilon rules and chain rules. (Those are rules of type $\alpha \to  \epsilon$ and $\alpha \to$ string of non-terminals only.) You can do this with every CF grammar. Since there's only a finite number of rules, let's denote $k$ the length of the longest right side of a rule from grammar $G$. Now take some word from L(G) and look at any of its syntax trees. On the first level of the syntax tree there is only one character - $\sigma$. On the sencond level there can be at most $k$ characters because that's the greatest number of characters we can get from one non-terminal. On level $h+1$ there can be at most $k^h$ characters so if you want to get a word of length greater than $k^h$ its syntax tree will have depth greater than $h+1$. However, the number of non-terminals in $G$ is also finite, let's say $n$. So if you ever wanted to get a word longer than $k^n$ it would have to have a syntax tree of depth greater than $n+1$. Now take a terminal from depth greater than $n+1$. It was derived from a non-terminal alpha, which was derived from a non-terminal beta, which wad derived from... which was derived from sigma. This sequence of non-terminals is of length greater than $n$ so there must be a non-terminal that appears in this sequence more than once. Now you can take this non-terminal and do exactly the same thing as you did with the repetitive states of DFA in the proof of pumping lemma for regular languages - you can either remove the part between the two apperances of the same non-terminal or add as many copies of it as you wish. And this is exactly what the lamma says. If there's anything you don't understand feel free to ask :).


The incircle of the triangle $ABC$ touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$.

Define the points $\{\begin{array}{c}X\in IC\cap DE\\\ Y\in EF\ ,\ IY\perp IC\\\ Z\in XY\cap ID\end{array}$. Prove that $CZ\perp BI$.

according again to this lemma from the begging of my proof from here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132338[/url], we observe that if $L \in BI \cap EF$, then $BL \perp CL$, so we reformulate the problem in the following way:

Let $ABC$ be a triangle, such that $L \in BI \cap EF$, and denote $Z \in DI \cap CL$, and $X,Y$ like in the initial problem. Prove that $X,Y,Z$ are collinear.

For this now we use only Menelaus , in triangle $EMD$, where $M \in ID \cap EF$, thus: $X,Y,Z$ are collinear $\Longleftrightarrow \frac{DZ}{ZM}={ID}{MI}$, which is true by some little computations on the line $ZD$ (we used that $\frac{EX}{XD}=1$, and $\frac{EY}{YM}=\frac{ID}{IM}$, because $ED \| IY$, because of the fact that $ED$ is the polar of $C$, thus $CI \perp DE$.)

$CI$ meets $EF$ at $K$. By applying the lemma that [b]pohoatza[/b] mention above we have $BK \bot CI$ $(1)$
We also have $KI/IX=KY/YE=ZY/YX$ so $ZK//IY$ but $IY\bot IC$ so $ZK\bot IC$ $(2)$
From (1) and (2) $B,K,Z$ are collinear .Then $CK\bot BZ$
Thus $I$ is the orthecenter of triangle $BZC$. This means $BI\bot ZC$ $q.e.d$


trigonometry
conics
ellipse
hyperbola
parabola
analytic geometry
Asymptote

What curve is defined by the polar equation
\[ p = \frac{1}{1 - \cos{\theta} + \sin{\theta}}? \]
(A) circle
(B) ellipse
(C) hyperbola
(D) parabola

$\rho = \frac{w}{1+e\cos(\theta-\phi)}$ is a conic

focus $O$
"directrice D"
excentricity $e$, $w=e.d(O;D)$
$\phi = angle( \overrightarrow{i},D) + \pi/2$

blahblahblah wrote:[hide]C)  It's not too hard to expand it and check the discriminant.  Anyone have an nicer solution?[/hide]

Just try  r-x+y = 1 or r = (1+x-y) or r^2 = (1+x-y)^2 = x^2+y^2. It's a  [hide]hyperbola  [/hide]

Also  notice that

 :sqrt: 2((sin(theta - pi/4) )  = sin(theta) - cos(theta) and you can write this in the standard form of  (a/r) = 1 + e Cos(theta- constant)  etc...

[quote="blahblahblah"]I don't have the general forms for conics in polar coordinates memorized, unfortunately.  Maybe I should derive them myself at some point.[/quote]

Deriving them for yourself once, especially after you have seen them once, is one of the absolute best ways to learn any math.

I agree.  Not only is it useful from a pedagogic point of view, it gives you the confidence that, no matter where you are, and no matter what texts or sources you have access to, that you can recreate the theorem or formula if need be.

In fact, that's usually how I study for tests, except that I also have to do a few plug and chug questions as well to make sure I don't do something like 1+1 = sqrt (2)

Hello, MCrawford!

[quote]What curve is defined by the polar equation:  $\rho = \frac{1}{1 - \cos{\theta} + \sin{\theta}}$ ?
(A) circle
(B) ellipse
(C) hyperbola
(D) parabola[/quote]
We have the conversions: $x = \rho\cos\theta,\;y = \rho\sin\theta,\;r^2 = x^2 + y^2$

Clear the denominator: $\rho - \rho\cos\theta + \rho\sin\theta \;= \;1 $

and we have: $\rho - x + y \;=\;1\;\;\;\Rightarrow\;\;\;\rho \;= \;1 + (x - y)$

Square both sides: $\rho^2 \;= \;1 + 2(x-y) + (x-y)^2$

Simplify: $x^2 + y^2 \;= \;1 + 2x - 2y + x^2 - 2xy + y^2\;\;\;\Rightarrow\;\;\;1 + 2x - 2y - 2xy \;= \;0 $

Solve for $y:\;\;y \;= \;\frac{2x + 1}{2(x + 1)} $

This is a [b](c) hyperbola[/b].
It has a vertical asymptote, $x = -1$, a horizotanl asymptote, $y = 1.$
Its two branches appear in the northeast and southwest "quadrants".


MATHCOUNTS
number theory
divisibility tests
prime numbers

Source: 2001 State Sprint

Given that m and n are digits, what is the sum of the values for m and n which yield the greatest six-digit number 5m562n that is divisible by 44?

[hide="my way"]I brutal forced it by trying devide 59562n by 44 and got lucky to have n = 8 so the answer is 17.  Any better way of doing this kind of problems?[/hide]

[quote="sheepwarrior"]you know 44 is didvisble by 4 and 11, so use divisibility tests for both.
for 11 you know that 5-m+5-6+2-n= a multiple of 11
you will end up having 6-m-n= something
since you are looking for largest m and n, see if -11 works
6-17=-11
so m=9 and n=8
and to test 4 (you already have number but try anyways) the last two digits should be multiple of four (28)[/quote]
I can't agree more with you, but it's much easier to start with the fours test first. Since that leaves you with n=0,4,or 8, you can easily implement these numbers into the test of 11s, starting with 8. Either way works though...

oohh.  I never knew there was a divisibility rule for 11.

thanks

There are at least 3 rules for divisibility by 11.

Take 2739

remove the last digit 9, then subtract 9 from the rest of the number 273

273 - 9 = 264

repeat the process by subtracting 4 from 26

26 - 4 = 22

Since 22 is divisible by 11, so is the original 2739.

That rule works largely the same way as the previous one.  You can also cut that last digit off multiply it by 10, add it to what's left of the number and repeat, but really, why would you :)

The rule for divisibily by 7, 13, and all other primes >5(i think) are similar.  For 7 you cut off the last digit, double it and subtract it from the rest of the number.

Take 392

Remove the 2, double it 2x2=4, subtract 4 from 39.  39 - 4 =35 which is div by 7, so the whole thing is divisible by 7!

Thanks frost... I didn't fully understand the seven rule until you explained it (again a credit to your good explaining skills, sorry this is somewhat a joke from another string)... and I think the alternate numbers rule is best for 11. What's the rule for 13?

The rule for 13 is to cut off the last digit, multiply it by 4, and add it to the rest of the number.  Or multiply it by 9 and subtract it from the rest.

Example 182

18 + 2(4) = 18 +8 = 26 which is divisible by 13

18 - 2(9) = 18 - 18 = 0 which is divisible by everything!

I am pretty sure that all prime numbers have two rules(one adding and one subtracting) that are similar to the ones I've given for 7, 11, and 13.  It is neat to notice that the two things you multiply by in the 13 rule add up to 13, while the two numbers you mutiply by in the 11 rule add up to 11...etc....

This was actually a Masters Round question from 1999 or so.

And I agree hockey, the alternating rule for 11 is much better.

Is 17 similar? Let's see, for 187, take off the last digit and multiply it by 3? 18 + 21 is 39. Might work.

How about 204? 20 + 12 is 32. Not this time.

By 8?

20 + 32 is 52, which works.

How about 187?

18 + 56 is 78, which also works!!!!

Since you add 8, you would subtract 9 times the units digit from the rest of the number.

20 - 36 = 16

Ummmm.....

[quote="236factorial"]Is 17 similar?[/quote]
Similar, yes.  But there is a different rule for each prime number.  Rules for 17 include:  Cut off the last digit, multiply it by 12, and add it to the rest of the number. Or multiply it by 5 and subtract it from the rest.

Informally, this rule works because 51 ("times 5, then subtract") and 119 ("times 12, then add") are multiples of 17.

The rules for 13 works because 39 ("times 4, then add") and 91 ("times 9, then subtract") are multiples of 13.

The rule for 7 works because 21 ("times 2, then subtract") is a multiple of 7.

If this was Master's Round material, it should be clear we are approaching the edge of MATHCOUNTS-level material.  So, let's not get too carried away, please.  For a formal explanation, please ask in one of the high school forums.

Interesting...
But just a question, how often does the divisibility rule for 13 17, etc useful?

Maybe seven is useful... and unless you master 7, 13, and 17 and become familiar with it (like for example 3, or 2) then it may be somewhat troublesome... but if you can learn it properly, then it's better than dividing out...

[quote="rcv"] But there is a different rule for each prime number. [/quote]

Are you sure. I remember that a lot of time ago (ten years or so) a friend and I tried to find that general rule, but we couldn't. We didn't have access to the web, then, after we were thought the rules for 7 and 11, we tried, and I think we found one for 13, 17, 23 and at least other two primes less than 100, but we really tried with many of them, and we failed. Of course, I was an 8th grader then, and I haven't try ever since, but I still don't think such a rule exists.

Best,

Here's a simplified version of what everyone has. I'll assume everyone knows 2,3,4,5,6,8,9,10,18,20,25...
Repeat all the steps as needed.

7: Last digit*2+Rest of the Number has to be a mult. of seven.
11: Best Way: Sum(even digits)-Sum(Odd digits)=Mult of 11 (including if its negative)
13: Last digit*4 + rest of the number is a multple of 13.
17: Last digit*12+Rest of the number

Alternate methods for 13,17: (the 17 one is easier than the top one)
13: Subtract (last digit*9) from the other two. The remainder has to be a mult of 13
17: Subtract (last digit*5) from the other two. The remainder has to be a mult of 17

actually, if no one noticed, my last post was a serious mistake. I assumed that 52 was divisible by 17, thus the solution not working.

For divisibility of 19, is the rule

If the units digits is doubled, the result is added to the rest of the digits, and the sum is divisible by 19, the number is divisible by 19.

What about 23? This is an unimportant dividing number.

oh, I just figured it out. The last digit times 7 added to the rest of the number has to be divisible by 23.

I that really true???

Didn't you already state the rule for 17? Multiply the units digit by 12 and then add it to the rest of the number. If the resulting number is divisible by 17, the number you started with is also divisible by 17.

The rule for 27 by the way is multiplying the units digit by 19 and apply the rest of the steps.

For 29, multiply the units digit by 3, and apply steps
For 31, multiply the units digit by 28, and apply steps
For 37, multiply the units digit by 26, and apply steps
For 41, multiply the units digit by 37, and apply steps...

Hopefully these are somewhat correct.
Do I need to go further?

I'd say just be familiar with 7,11,13,15 and (optional)17. 23 and above is nice, but not necessary...

[quote="rcv"] But there is a different rule for each prime number. [/quote]

[list]
Wisdom is something we get when it is not useful anymore.
[b]Gabriel Garcia Marquez[/b]
[/list]

Of course such rules exists for any prime different from 2 and 5. The key is just to find $10^{-1} \mod p$, what is always possible (that is, $10^{-1}$ we will define it as the smallest possitive number $k$ such that $10k\cong 1 \mod p$), then, you multiply the last digit by $k$ and add it to the other part, and go by steps. It always work. To find $k$ is not that hard, you can apply the generalized euclid algorithm.

Best,


1. Express in simplest form the numerical value of $\frac{\sin 40^\circ+\sin 80^\circ}{\cos 40^\circ+\cos 80^\circ}$.

2. If $2^{22}=4,194,304$ find the sum

$1*2+2*2^2+3*2^3+...+21*2^{21}=\displaystyle\sum_{n=1}^{21}{n*2^n}$

[hide="hint for 2"]
Use the same method that you use for a simple geometric series.
[/hide]

[quote="citistarz"]Isn't $\frac{sin60}{cos60}=tan60=\sqrt3$?[/quote]

And in what way is that different than dividing $\sin 60=\sqrt{3}/2$ and $\cos 60=1/2$ out?

Yeah, and who memorizes $\tan$ anyway? :P I always just do $\sin \theta/\cos\theta$. ;)

You dont need to memorize tan  :P you can figure it out most of the time.


Let  $ A \in S_{n} \mathbb{(R)}$,
$ I_{n}$ is identity nxn matrix.
[b]Such that :[/b]
$ e^{A}=I_{n}$ then $ A=O_{n}$
 :roll:

$ A$ is diagonalizable. What are its eigenvalues?

I would be more correct to write $ A\in O_n$

It's a notation conflict. I think QuyBac really means the zero matrix, not the orthogonal group.

I would just call the zero matrix $ 0$- the number, not the letter. It doesn't even need a size marker.

My other notation question: what does he mean by $ S_n(\mathbb{R})?$

[quote="jmerry"]From context, the symmetric matrices.[/quote]
yes ! symmetric matrices ! :oops:


pigeonhole principle
calculus
calculus computations

consider the sequence:  $ \frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},...$.
show that for any $ a\in [0,1]$, there is a subsequence that converges to $ a$.

Using pigeonhole principle, you can always find $ 0 < k_n < n$ suth that $ \left| a \minus{} \frac{k_n}{n} \right| \leq \frac{1}{n}$ for any $ n \geq 2$.

Then just choose a subseqence $ \frac{k_2}{2}, \frac{k_3}{3}, \frac{k_4}{4}, \cdots$.

[quote="sos440"]Using pigeonhole principle, you can always find $ 0 < k_n < n$ suth that $ \left| a \minus{} \frac {k_n}{n} \right| \leq \frac {1}{n}$ for any $ n \geq 2$.
[/quote]

can you prove this?


analytic geometry
geometry
USAMTS
arithmetic sequence
probability
expected value
complex numbers

Hi My teacher assigned me with these problems but i cant manage to solve them so can anyone help me with these?

Problem 1
An increasing arithmetic sequence with infinitely many terms is determined as follows.
A single die is thrown and the number that appears is taken as the first term. The die is thrown again and the second number that appears is taken as the common difference between each pair of consecutive terms. Determine with proof how many of the 36 possible sequences formed in this way contain at least one perfect square.

Problem 2
George has six ropes. He chooses two of the twelve loose ends at random (possibly from the same rope), and ties them together, leaving ten loose ends. He again chooses two loose ends at random and joins them, and so on, until there are no loose ends. Find, with proof, the expected value of the number of loops George ends up with.

Problem 3
Let r be a nonzero real number. The values of z which satisfy the equation
R^4z^4 + (10r^6 - 2r^2)z^2 - 16r^5z + (9r^8 + 10r^4 + 1) = 0 are plotted on the complex plane (i.e. using the real part of each root as the x-coordinate and the imaginary part as the y-coordinate). Show that the area of the convex quadrilateral with these points as vertices is independent of r, and find this area.

Problem 4
Homer gives mathematicians Patty and Selma each a di_erent integer, not known to the other or to you. Homer tells them, within each other’s hearing, that the number given to Patty is the product ab of the positive integers a and b, and that the number given to Selma is the sum a + b of the same numbers a and b, where b > a > 1. He doesn’t, however, tell Patty or Selma the numbers a and b. The following (honest) conversation then takes place:
Patty: “I can’t tell what numbers a and b are.”
Selma: “I knew before that you couldn’t tell.”
Patty: “In that case, I now know what a and b are.”
Selma: “Now I also know what a and b are.”
Supposing that Homer tells you (but neither Patty nor Selma) that neither a nor b is greater
than 20, find a and b, and prove your answer can result in the conversation above.

Problem 5
Given triangle ABC, let M be the midpoint of side AB and N be the midpoint of
side AC. A circle is inscribed inside quadrilateral NMBC, tangent to all four sides, and that circle touches MN at point X. The circle inscribed in triangle AMN touches MN at point Y , with Y between X and N. If XY = 1 and BC = 12, find, with proof, the lengths of the sides AB and AC.

STOP posting these.

It's the third time you're posting them on the forum. Don't do that.

And do you really think we wouldn't recognize the current USAMTS problems?

Topic locked.


superior algebra
superior algebra unsolved

This [b]post[/b] has to do with [b]wiki[/b]'s entry on [b]Baer's sum[/b]...

[b][color=red]Background.[/color][/b] Given two extensions

$ 0 \to B \to E \to A \to 0$

and

$ 0 \to B \to F \to A \to 0$

their [b]Baer sum[/b] is constructed as follows:

a) First off, we form the pullback $ K$ of the morphisms $ E \to A$ and $ F \to A$.

b) Now, if we let $ Z$ denote the [b]quotient[/b] $ K/\Delta$ where $ \Delta \equal{} \{( \minus{} b,b) : b \in B\}$, then the sequence

$ 0 \to B \to Z \to A \to 0$

is the [b]Baer sum[/b] of extensions 1 and 2 above.

[b][color=red]-----     -----    -----     -----     -----     -----[/color][/b]

What [b]I don't[/b] understand is how it is that we're supposed to [i]regard[/i] the $ \Delta$ over there.

In order to form the quotient $ K/\Delta$ we must have first $ K \supseteq \Delta$, and this is not always the case here, right? How are we supposed to sort the anomaly in this ill-defined version of [b]Baer's sum[/b]?

Can any of you guys shed some light on this mystery?  :wink:

The notations are a bit dirty.

What is meant is the following: Given two extensions
$ 0\to B\overset{e}{\to} E\overset{e^{\prime}}{\to} A\to 0$
and
$ 0\to B\overset{f}{\to} F\overset{f^{\prime}}{\to} A\to 0$,
we set
$ K \equal{} \left\{\left(x,y\right)\in E\times F\mid e^{\prime}\left(x\right) \equal{} f^{\prime}\left(y\right)\right\}\subseteq E\times F$
and
$ \Delta \equal{} \left\{\left( \minus{} e\left(b\right),f\left(b\right)\right)\mid b\in B\right\}\subseteq E\times F$.
Then, $ \Delta\subseteq K$ can be easily seen.

Since $ e$ and $ f$ are injections, people often write $ b$ for $ e\left(b\right)$ or $ f\left(b\right)$, but I wouldn't recommend doing this unless one is completely aware of what one is doing.

  darij

I see...  :lol:

Thanks a lot for taking the time to reply, dear Sir. Your explanation was very helpful to me.


Let $ x,\ y$ be real numbers such that $ x+y=3,\ xy=-2$. Find the values of $ x^{2}+y^{2},\ x^{3}+y^{3}$ and $ x^{5}+y^{5}$.

[hide="Not 100% Confident"]$ x^{5}+y^{5}= (x^{2}+y^{2})(x^{3}+y^{3})-(x^{2}y^{3}+y^{2}x^{3})$ You know that $ x^{2}y^{3}+y^{2}x^{3}$ can be expressed as $ (xy)(xy)(y)+(yx)(yx)(x)$, which equals $ 4y+4x$. Now, you can factor out the 4, giving you $ 4(x+y)$, which equals $ 12$. If you did the previous problems, you know that $ (x^{2}+y^{2})(x^{3}+y^{3})$ equals $ 13*45$, which equals $ 585$. Thus, the solution that I get to this problem is $ 585-12=573$.  [/hide]


Determine all trple of real numbers $ x;y;z$ such that
$ x^6\plus{}y^6\plus{}z^6\minus{}6(x^4\plus{}y^4\plus{}z^4)\plus{}10(x^2\plus{}y^2\plus{}z^2)\minus{}2(x^3y\plus{}y^3z\plus{}z^3x)\plus{}6(xy\plus{}yz\plus{}xz)\equal{}0$

[quote="negative"]Determine all trple of real numbers $ x;y;z$ such that
$ x^6 + y^6 + z^6 - 6(x^4 + y^4 + z^4) + 10(x^2 + y^2 + z^2) - 2(x^3y + y^3z + z^3x) + 6(xy + yz + xz) = 0$[/quote]
[hide]$ \sum_{cyc}(x^6 - 6x^4 + 10x^2 - 2x^3y + 6xy) = \sum_{cyc}(x^3 - 3x - y)^2.$
But $ x^3 - 3x = y\Leftrightarrow(x + 2)(x - 1)^2 = y + 2\Leftrightarrow(x - 2)(x + 1)^2 = y - 2.$
It gives $ \{x,y,z\}\subset[ - 2,2].$
Thus, we can assume $ x = 2\cos\alpha,$ $ y = 2\cos\beta$ and $ z = 2\cos\gamma,$ which gives
$ 3\alpha = \pm\beta + 2\pi k,$ $ 3\beta = \pm\gamma + 2\pi n$ and $ 3\gamma = \pm\alpha + 2\pi m,$ where $ \{k,l,m\}\subset\mathbb Z,$
which easy to check.
For example, $ 3\alpha = \beta,$ $ 3\beta = \gamma$ and $ 3\gamma = \alpha + 2\pi$ gives $ \left(2\cos{\frac {\pi}{13},2\cos{\frac {3\pi}{13},2\cos{\frac {9\pi}{13}\right);}}}$
$ 3\alpha = \beta,$ $ 3\beta = \gamma$ and $ 3\gamma = -\alpha + 4\pi$ gives $ \left(2\cos{\frac {\pi}{7},2\cos{\frac {3\pi}{7},2\cos{\frac {5\pi}{7}\right).}}}$ :wink:
[/hide]


Let $ a,b,c$ be positive reals, show that
\[ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2 \geq abc(a^2 \plus{} b^2 \plus{} c^2)\].

I tried it using weighted AM-GM but did not succeed.

wrong again! mods if posiible delete this..

this inequality is true -
$ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2 \ge a^2b^2c \plus{} b^2c^2a \plus{} c^2a^2b$

[quote="MayankM"]Let $ a,b,c$ be positive reals, show that
\[ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2 \geq abc(a^2 \plus{} b^2 \plus{} c^2)\]
.

I tried it using weighted AM-GM but did not succeed.[/quote]

the inequality does not hold... :wink:

try $ a\equal{}1,b\equal{}2,c\equal{}10$

Thanks kuing...saved me considerable effort.


is something missing?
is there some relation among $x_1,x_2,\cdots,x_n$?

It suffices to prove it just for $n=1$ (then, just add the obtained inequalities for $x_i$, $i=1\ldots n$), that is \[ 5\sqrt[5]{x^3}-3\sqrt[3]{(\frac{3x+2}{5})^5}\leq 2 \] or \[ \sqrt[5]{x^3}\leq \cfrac{2+3\sqrt[3]{(\cfrac{3x+2}{5})^5}}{5} \]
The RHS due to cauchy is bigger than \[ \sqrt[5]{1 \cdot 1 \cdot \sqrt[3]{(\frac{3x+2}{5})^5} \cdot \sqrt[3]{(\frac{3x+2}{5})^5} \cdot \sqrt[3]{(\frac{3x+2}{5})^5}} \] which is equal to  \[ \frac{3x+2}{5} \] which, again by cauchy, is bigger than \[ \sqrt[5]{1\cdot1\cdot x \cdot x \cdot x}=\sqrt[5]{x^3}. \]  :roll:  Equality for $x=1$.


Find the smallest prime number $p$ that cannot be represented in the form $|3^{a} - 2^{b}|$, where $a$ and $b$ are non-negative integers.

I think it is $17$, however I havent checked all the cases, what one should try is to look what pairs $(a,b)$ modulo 16, give $3^a-2^b$ a multiple of 17, and show using factorizations, modulo 5, modulo 11, modulo 16, or more, to show they have more factors or the equivalence is not truth!

17=81-64...

however, it looks like 37 does not work. The idea to prove it should be the same, still.

[quote="da-kid"]I think it's 3. For it to be 3, 2^b would need to be a multiple of 3. You can factor out a 3 from 3^a making it 3^(a-1) * 3. To get 3, 2^b would need to become (3^(a-1) +/-1) * 3 and because 2^b is not a multiple of 3, this could never occur.[/quote]
Be careful! The nonnegative integers include 0, so $ 3\equal{}|3^{0}\minus{}2^{2}|$ can be represented.

[quote="FINNN"]Hey mathtastic, now that you no longer have a nice list of solutions for $p<41$, I see you've "made a claim". Unfortunately, if you were to submit this proof for an olympiad asking "what is the smallest", you would get perhaps 50% of the possible points. Proving something to be a solution is quite trivial compared to showing that it is the smallest. You'll have to rethink your proof.[/quote]
Why don't you try the problem yourself? AkshajK just showed how simple it is to make all of the smaller primes. With 41 you get stuck so make a claim and prove it. It isn't trivial to prove but it is trivial to verify all smaller primes work.

Notice that

$b=5$
$a=2(log_3(i)+1)$

proves that $41$ is indeed possible!

a and b are non-negative integers, not complex numbers.

[quote=AkshajK]Actually it is pretty easy to show that it is the smallest because \begin{eqnarray*} 37&=& 64 - 27\\31&=& 32 - 1\\29&=& 32 - 3\\23&=& 32 - 9\\19&=& 27 - 8\\17&=& 81 - 64\\13&=& 16 - 3\\11&=& 27-16\\7&=& 16-9\\5&=& 9-4\\3&=& 4-1\\2&=& 3-1 \end{eqnarray*}[/quote]

Hi!  :)

We need to show that 41 is the smallest value of $p$ satisfying the condition.

Suppose that $|3^a-2^b|=41$

So we have two cases:

1) $3^a-2^b=-41$.

\begin{align*}
2^b-3^a&=41\\
2^b-3^a&=2^5+3^2\\
2^b-2^5&=3^a+3^2>0
\end{align*}

So, $b>5$. That implies $32|3^a+3^2$. Thus $8|3^a+3^2$, wich is equivalent to $3^a+3^2\equiv 0\mod 8$. But

\begin{align*}
3^a+3^2&\equiv 0\mod 8\\
\iff 3^a&\equiv -3^2\mod 8\\
\iff 3^a&\equiv 7 \mod 8
\end{align*}

And the last equation is false, because $3^{a}\equiv 1\mod 8\lor 3^{a}\equiv 3\mod 8$.

2)$3^a-2^b=41$

\begin{align*}
3^a-2^b&=41\\
3^a-2^b&=2^5+3^2\\
3^a-3^2&=2^b+2^5>0
\end{align*}

So, $a>2$. That implies $9|2^b+2^5$, wich is equivalent to $2^b+2^5\equiv 0\mod 9$. It's easy to see that this only happens when $b=6k+2$ for some integer $k$ (because $2^6\equiv 1\mod 9$). Finally:

\begin{align*}
3^a-3^2&\equiv 2^5+2^b \mod 13\\
\iff 3^a&\equiv 3^2+2^5+\left(2^6\right)^k 2^2\mod 13\\
\iff 3^a&\equiv 9+6+\left(1\right)^k 4\mod 13\\
\iff 3^a&\equiv 6\mod 13
\end{align*}

And the last equation is false, because $3^{a}\equiv 1\mod 13\lor 3^{a}\equiv 3\mod 13\lor3^{a}\equiv 9\mod 13$.

Sorry if my english is not perfect.... Please tell me if I am wrong.

[quote=mathtastic][hide="Mod Bash"]
We claim that the smallest prime that cannot be written in the given form is $41$. It's easy to verify that all primes under $41$ can be written in the given form.
[list][*][b]Case 1:[/b] $3^a-2^b=41$.

It is clear that $a\ge 1$, so we can safely say that $3^a\equiv 0\pmod{3}$. From here we can take $\pmod{3}$ to get that $-2^b\equiv 41\equiv 2\pmod{3}\implies 2^b\equiv 1\pmod{3}$. Checking values of $2^b\pmod{3}$ we see that $2^b\equiv 1\pmod{3}\iff b\equiv 0\pmod{2}$. Thus we can set $b=2k\implies 2^b=2^{2k}=(2^k)^2$. Let's leave this aside for now and go back to our original equation. It is clear that $b\ge 2$, so we can safely say that $2^b\equiv 0\pmod{4}$. From here we can take $\pmod{4}$ to get that $3^a\equiv 41\equiv 1\pmod{4}$. Checking values of $3^a\pmod{4}$ we see that $3^a\equiv 1\pmod{4}\iff a\equiv 0\pmod{2}$. Thus we can set $a=2j\implies 3^a=3^{2j}=(3^j)^2$. Now we can see where we're getting. We have $(3^j)^2-(2^k)^2=41\implies (3^j-2^k)(3^j+2^k)=41$. Clearly, since $41$ is prime and $3^j+2^k\not=1$, we must have $3^j-2^k=1$ and $3^j+2^k=41$. Adding these equations and dividing by $2$ gives $3^j=21$ which clearly isn't possible. Thus this case has no solutions. $\square$

[*] [b]Case 2:[/b] $2^b-3^a=41$.

It is clear that $a\ge 1$, so we can safely say that $3^a\equiv 0\pmod{3}$. From here we can take $\pmod{3}$ to get that $2^b\equiv 41\equiv 2\pmod{3}$. Checking values of $2^b \pmod{3}$ we see that $2^b\equiv 2\pmod{3}\iff b\equiv 1\pmod{2}$. This will be important later, so keep it in mind. It is clear that $b\ge 2$, so we can safely say that $2^b\equiv 0\pmod{4}$. From here we can take $\pmod{4}$ to get that $-3^a\equiv 41\equiv 1\pmod{4}\implies 3^a\equiv 3\pmod{4}$. Checking values of $3^a\pmod{4}$ we see that $3^a\equiv 3\pmod{4}\iff a\equiv 1\pmod{2}$. Since we still aren't done, let's try another mod. Let's try $8$. We have $2^b-3^a\equiv 1\pmod{8}$. Values of $2^b\pmod{8}$ are $1,2,4,0,0,0,\cdots$. Values of $3^a\pmod{8}$ are $1,3,1,3,1,3,\cdots$. However, since we know that $a\equiv 1\pmod{2}$ we [b]must[/b] have $3^a\equiv 3\pmod{8}$. Thus we now have $2^a-3\equiv 1\pmod{8}\implies 2^a\equiv 4\pmod{8}\implies a=2$. However, $a=2$ won't work because we have $a\equiv 1\pmod{2}$. Thus this case also has no solutions. $\square$[/list]

Since we have exhausted both cases, we are done (whew!)
$\mathbb{Q}.\mathbb{E}.\mathbb{D}.$
 [/hide][/quote]

Amazing solution!  :trampoline:

"Try this problem, it's a better one out of the oly problems on the handout"- Math Circle
*Some time later*
"I finished the problem, can you go over it at the end"-me
"Lol you did that dumb problem"- Math Circle

[hide=Sol for storage]
We can show that all of the primes below $41$ are possible.
$2=|3^1-2^0|$
$3=|3^0-2^2|$
$5=|3^1-2^2|$
$7=|3^0-2^3|$
$11=|3^3-2^4|$
$13=|3^1-2^4|$
$17=|3^4-2^6|$
$19=|3^3-2^3|$
$23=|3^3-2^2|$
$29=|3^1-2^5|$
$31=|3^0-2^5|$
$37=|3^3-2^6|$

Now we can show that $41$ is the answer.

Case 1: $2^b-3^a=41$
$3^a\equiv 1,3\mod 8$, so $2^b\equiv 2,4\mod 8$. This means $b=1,2$, which is impossible.

Case 2: $3^a-2^b=41$
$3^a\equiv 1,3\mod 8$, so $2^b\equiv 0,2\mod 8$. $b=1$ doesn't work, so $2^b\equiv 0\mod 8$ and $3^a\equiv 1\mod 8$. Thus $a$ is even. $2^b\equiv 1\mod 3$, so $b$ is also even. Then let $3^a=x^2$ and $2^b=y^2$. If $x^2-y^2=41$, then $(x-y)(x+y)=41$, so $x=21,y=20$. Since $21$ and $20$ are not powers of $3,2$, there are no solutions for $a,b$, as desired.
[/hide]


Me again  :D

[hide]25*6=8*x
x=18.75

you need 19 people [/hide]

[hide]
8X=150
X=150/8=18.75

You need 19 OR MORE people (more people to get it done less than 8 hours).
[/hide]

[hide][quote="aznmonkey1992"]a job  can be done in 25 hours by 6 poeple.
how many people are needed to get it done in 8 hrs or less?[/quote]
If 6 people can do it in 25 hours, then 1 person can do it in 150 hours.
to do in 8 hours you need 150/8=18.75 people but since you can't have 3/4ths of a person so you need a minimum of 19people but you can also have more people.[/hide]


quadratics
factorial
algebra
function
domain
Ring Theory
superior algebra

For an element in a quadratic field, does irreducibility imply primality and vice versa?

Irreducibility only implies primality in factorial domains; in fact a factorisation into irreducibles is unique iff the irreducibles are prime. So you will find counterexamples in any nonfactorial quadratic field.

Alas, both statements are false. First, there are non-UFDs where irreducible $\implies$ prime, e.g. this is vacuously true in rings with no irreducibles, such as the ring of all algebraic integers, where every elt factors $x = {\sqrt x}{\sqrt x}$. Also the direction $(\implies)$ in the second statement is false. Can you think of an interesting counterexample?


Given $ a, b, c, d \ge 1$ satisfy $ a\plus{}b\plus{}c\plus{}d\equal{}8$. Prove that:
\[ \frac{1}{1\plus{}a^3}\plus{}\frac{1}{1\plus{}b^3}\plus{}\frac{1}{1\plus{}c^3}\plus{}\frac{1}{1\plus{}d^3} \ge \frac{4}{9}\]
 :)

[quote="nguoivn"]Given $ a, b, c, d \ge 1$ satisfy $ a \plus{} b \plus{} c \plus{} d \equal{} 8$. Prove that:
\[ \frac {1}{1 \plus{} a^3} \plus{} \frac {1}{1 \plus{} b^3} \plus{} \frac {1}{1 \plus{} c^3} \plus{} \frac {1}{1 \plus{} d^3} \ge \frac {4}{9}\]
:)[/quote]
it seems very easy using the well known fact that for $ x,y\ge 1$,
\[ \frac1{1 \plus{} x^2} \plus{} \frac1{1 \plus{} y^2}\ge\frac2{1 \plus{} xy}\]
thus $ \frac {1}{1 \plus{} a^3} \plus{} \frac {1}{1 \plus{} b^3} \plus{} \frac {1}{1 \plus{} c^3} \plus{} \frac {1}{1 \plus{} d^3} \ge \frac {4}{1 \plus{} (abcd)^\frac34}\ge \frac {4}{9}$

 Trivially it holds for all $ a, b, c, d > \minus{} 1$ satisfying $ a \plus{} b \plus{} c \plus{} d \equal{} 8$: supposing e.g. $ \minus{} 1< a\le 1, \frac {1}{1 \plus{} a^3}\ge \frac 12 > \frac {4}{9}$ :)

[quote="nguoivn"]Given $ a, b, c, d \ge 1$ satisfy $ a \plus{} b \plus{} c \plus{} d \equal{} 8$. Prove that:
\[ \frac {1}{1 \plus{} a^3} \plus{} \frac {1}{1 \plus{} b^3} \plus{} \frac {1}{1 \plus{} c^3} \plus{} \frac {1}{1 \plus{} d^3} \ge \frac {4}{9}\]
:)[/quote]

$ f(x)\equal{}\frac{1}{1\plus{}x^3} \ge \frac{11}{27}\minus{}\frac{4x}{27}$    for $ x >\minus{}1$;

$ f(a)\plus{}f(b)\plus{}f(c)\plus{}f(d) \ge \frac{44}{27}\minus{}\frac{4}{27}(a\plus{}b\plus{}c\plus{}d)\equal{} \frac{4}{9}$

Equality holdes for : $ a\equal{}b\equal{}c\equal{}d \equal{} 2$;

[quote="nkp"][quote="Sergic Primazon"]
$ f(x) \equal{} \frac {1}{1 \plus{} x^3} \ge \frac {11}{27} \minus{} \frac {4x}{27}$    for $ x > \minus{} 1$;
[/quote]
Can you explain me how do you arrive to this inequality ?[/quote]

It rewrites as :

\[ \frac{(x\minus{}2)^2(4x^2\plus{}5x\plus{}4)}{27(x\plus{}1)(x^2\minus{}x\plus{}1)}\ge 0\]

;)

[quote="nkp"][quote="Sergic Primazon"]
$ f(x) \equal{} \frac {1}{1 \plus{} x^3} \ge \frac {11}{27} \minus{} \frac {4x}{27}$    for $ x > \minus{} 1$;
[/quote]here
Can you explain me how do you arrive to this inequality ?[/quote]
this method is called "tangent line method" see the attached file [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=645777]here[/url]


Yeah, I posted a similar thread in the chem section. You know sometimes I wonder if kids should start early and dream big....

My younger sister in middle school is interested in participating in the biology olympiad in high school. Do you know of any summer programs or other educational opportunities that would be good for learning basic and collegiate level bio?

Yes, I actually partook in a middle school biology course, 1 year ago in my rising 6th grade summer. It was quite interesting.

Try CTY- Biology of the senses class.

First, you must take a pre-evaluation exam, of your arithmetic and analogy skills, to allow you to enter the course. Then you are free to pick from various CTY class including Crystals and Polymers (chemistry), Biology of the Senses (the one I'm recommending), and many others.

The class I recommend, Biology of the senses, takes a further look unto the nervous system, and sensual organs of that system. There highlights of the course include: learning about the functions of certain nerves, dissecting squid, cow's eyes, and cat brains, and delivering a final presentation on the nervous system of any animal of your choosing.

It's great fun, and a course which recquires a good power of rote memorization. It is a good start for any aspiring neurologist or biologist.

Campuses vary from stay-in for out of state participators in which students stay at dorms, and local, in which students are picked up at the end of the day by their parents.

Elligibility for this particular course ranges from rising 6th to rising 7th.

I hope i have helped.

-arb95


I would like to know if it's possible to prove algebraically the following equality.

$ \left(\sqrt{1\plus{}(\tan 72^{\circ})^2} \minus{} 1 \right)^2 \equal{} 5$

Thanks in advance.

hello, you will need a formula for $ \tan(72^{\circ})$ by radicals, use the addition formuals to evaluate $ \tan(5x)$  and set $ x \equal{} 72^{\circ}$.
Sonnhard.

Use $ \sqrt{tan^2 x \plus{} 1} \equal{} sec x$ and $ cos 72 \equal{} \frac{\sqrt{5} \minus{} 1}{4}$

Thank you to both Dr Sonnhard Graubner and qwerty414 for your quick answer.

Following your suggestions, I proved the equality. However, in a olympiad I think it wouldn't be practical to find $ \cos(72^{\circ})$. Probably, I will have to keep in mind more values beyond the standard ones...

Thinkertoys


Hi every one,

I have a revenue function R(x) , Cost function C(x), so I can calculate the profit function P(x)=R(x)-C(x).

But, how can I solve the following questions:

1. At what level of production is the company profitable?

2. Find the total profit at production levels at which the company is profitable.

THanks in Advance.

Are you missing some information regarding the numbers, or am I being ignorant of this? Apparently this is supposed to be some type of optimization (maxima/minima) problem, right?

[quote="Tyl"]Being profitable at some level of production means for that $ x$ we have $ P(x) > 0$. So find those $ x$.

Don't really understand the second question very well.[/quote]

Dear Tyl : I also understand that we need to find those x such as $ P(x)>0$. but how can we use maxima to find those $ x$.

Dear gauss1181 : The previous questions are to find the maximum profit. (It was easy).

The point is that I need to use maxima to find the solutions.

Thanks in advance.


$ \begin {cases} \sqrt {\frac{x\plus{}1}{2y\minus{}1}\plus{}\sqrt {\frac{2y\minus{}1}{x\plus{}1}}}\equal{}\frac{5}{2} \\ x\minus{}y\equal{}2 \end {cases}$

Perhaps it could work like this:
$ x\minus{}y\equal{}2 <\equal{}> x\plus{}1\equal{}y\plus{}3$

If we insert this in the first equation, then we have:
$ \frac{y\plus{}3}{2y\minus{}1} \plus{}\sqrt{\frac{2y\minus{}1}{y\plus{}3}}\equal{}6,25$

We substitute: $ \frac{y\plus{}3}{2y\minus{}1}\equal{}z^2$

$ z^2\plus{}1/z\equal{}6,25 <\equal{}>z^3\plus{}1\equal{}6,25z <\equal{}> z^3\minus{}6,25z\plus{}1\equal{}0$

This equation can be solved and then you have to transform the solutions with a re-substitution.

[quote="Atan"]Perhaps it could work like this:
$ x \minus{} y \equal{} 2 < \equal{} > x \plus{} 1 \equal{} y \plus{} 3$

If we insert this in the first equation, then we have:
$ \frac {y \plus{} 3}{2y \minus{} 1} \plus{} \sqrt {\frac {2y \minus{} 1}{y \plus{} 3}} \equal{} 6,25$

We substitute: $ \frac {y \plus{} 3}{2y \minus{} 1} \equal{} z^2$

$ z^2 \plus{} 1/z \equal{} 6,25 < \equal{} > z^3 \plus{} 1 \equal{} 6,25z < \equal{} > z^3 \minus{} 6,25z \plus{} 1 \equal{} 0$

This equation can be solved and then you have to transform the solutions with a re-substitution.[/quote]

[color=darkblue]Thank you very much. I try to factor equation $ x^3 \minus{} 6,25z \plus{} 1 \equal{} 0$, but I can't.[/color]

Yeah, it seems to be, that this cubic equation has no simple solutions, although they all are real numbers. I hope, you know, how to solve a cubic equation with the help of Cardano's formulas, so that you receive the exakt solutions.

Otherwise, I have found an outline, how to cope with that equations, but it's with german language. So I can translate a little bit the written words:
In the headline, there stand: "with real r, s and t and complex x"
Then, there are three cases depending on the number D:
-left case: D<0 -> three different real solutions
-middle case: D=0 -> three real solutions, but two are identical
-right case: D>0 -> one real and two complex solutions

You will need to follow the left path.


Chuck a luck is a game in medieval fairs where players play one dollar to play.  He then picks a number from 1-6 and rolls three dice.  For every time that he gets his number rolled, he gets a dollar.  What is the average gain.  (Let loss be negative gain.)

[hide]i'm not sure what i did but i got 0[/hide]

Wrong.

EDIT: Misread it >_> . Solution is null and void.

[hide="Solution"]
The chance of [b]not[/b] getting your number turned up on three rolls is:

(5/6) :^3:
(125/216)

125/216 of the time you win, so 91/216 of the time you lose.

-34/216 = [b]-17/108, or a little less than 17 cents per game.[/b]
[/hide]

hmm im getting something different...

[hide]
the total possible results of the game is that he ends up with -1, 0, +1, or +2.  the average gain is:

Average Gain = -1*(probability of -1)  + 0*(probability of 0) + 1*(probability of 1) + 2*(probability of 2)
= -1*(5/6)^3      +       0        +      1*(3c2)*(1/6)^2*(5/6)      +     2*(1/6)^3
= -.50

hmm nice answer (on average losing 50 cents per play) ... but is the solution nice :)?[/hide]

[quote="Treething"]Wrong.

EDIT: Misread it >_> . Solution is null and void.

[hide="Solution"]
The chance of [b]not[/b] getting your number turned up on three rolls is:

(5/6) :^3:
(125/216)

125/216 of the time you win, so 91/216 of the time you lose.

-34/216 = [b]-17/108, or a little less than 17 cents per game.[/b]
But he gets a dollar for each time it shows up, not if it just shows up
[/hide][/quote]
0 wins 125/216
1 win 75/216
2 win 15/216
3 wins 1/216

So 125/216*-1+75/216*0+15/216*1+1/216*2=-108/216=-1/2


A certain calculator has only two keys $ [\plus{}1]$ and $ [\times 2]$. When you
press one of the keys, the calculator automatically displays the result.
For instance, if the calculator originally displayed `9' and you pressed
$ [\plus{}1]$, it would display `10.' If you then pressed $ [\times 2]$, it would
display `20.' Starting with the display `1,' what is the fewest number
of keystrokes you would need to reach `200'?

umm_
the trick is to get a number where, when multiplied by 2 a few times, will give you 200. you want it as small as possible

so_
200/2/2/2=25
cant get smaller_
so we try to get 25-1 using 2s

24/2/2/2=3
thats as small as it gets_

so from one_
X2
+1
X2
X2
X2
+1
X2
X2
X2
=200

smallest is 9


[quote="jlebian"]The game of Rock Paper Scissors is played by three players. Each player begins by making a fist; then, on the count of four, shows either four fingers (paper), two fingers (scissors) or a fist (rock). What is the probability that exactly two players show the same sign? Express your answer as a common fraction.[/quote]

[hide]3/3*1/3*2/3*3= 2/3[/hide] Maybe I did something wrong... :|

I got[hide]2/9 since all three can't be the same[/hide]

[quote="ch1n353ch3s54a1l"][hide]I got $\frac{2}{3}$ too...
Number of arrangements all get different: $3!=6$
Number of arrangements all get same: $3$
Number of total arrangements: $27$
$1-\displaystyle\left(\frac{6}{27}+\frac{3}{27}\right)=\boxed{\frac{2}{3}}$[/hide][/quote]
Ah, ok, that makes sense. Thanks. =D
Dogseatcheese, your answer was right but I didn't get how you got it.
 :P

[hide]

Let's assume that two go scissors and one goes something else.

The probability of the first guy goes scissors is $1/3$, second guy going scissors is $1/3$, third guy NOT going scissors is $2/3$. That leaves with with $2/27$. Now, you know that there are three combinations of two people, not just one with the first guy and secnd guy. So, you multiply by three to get $2/29$ Also, paper and rock works too, so you multiply by three again to take into account 3 different signs, not just one. The final answer is $2/3$. $3*3*1/3*1/3*2/3$[/hide]

[hide]
There are 6 ways for them to be all different and 3 for them to all be the same.
Thus 9 ways that don't work.
$3^3=27$ total, so $27-9-18$ working ways
$18/27=\frac 23$[/hide]

[hide]3! ways to get different, 3 ways to get same, 27-6-3=18, $\frac{18}{27}=\frac{2}{3}$[/hide]

[hide]i got 1/2, trial and error. am i doing it right?[/hide]

edit: sorry, i forgot about [hide]them all getting the same. now i get 2/3.[/hide]

[quote="Klebian"]The game of Rock Paper Scissors is played by three players. Each player begins by making a fist; then, on the count of four, shows either four fingers (paper), two fingers (scissors) or a fist (rock). What is the probability that exactly two players show the same sign? Express your answer as a common fraction.[/quote]

[hide]
27 total possiblilities.
6 all different
3 all same
Therefore there are 18 ways
18/27=2/3[/hide]

Do I have to show my work? It's the same as anyone elses
The answer is 2/3


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The following is the second problem from the first round of the greek competition.

Let $ f: \mathbb{R}\longrightarrow \mathbb{R}$ be a function and it holds that $ f\left(x\minus{}f\left(y\right)\right)\minus{}f\left(y\minus{}f\left(x\right)\right)\equal{}2f\left(f\left(x\right)\minus{}f\left(y\right)\right), \forall x,y\in \mathbb{R}$. Prove that $ f\left(x\minus{}f\left(x\right)\right)\equal{}0,\forall x\in \mathbb{R}$

[quote="Γιώργος"]The following is the second problem from the first round of the greek competition.

Let $ f: \mathbb{R}\longrightarrow \mathbb{R}$ be a function and it holds that $ f\left(x \minus{} f\left(y\right)\right) \minus{} f\left(y \minus{} f\left(x\right)\right) \equal{} 2f\left(f\left(x\right) \minus{} f\left(y\right)\right), \forall x,y\in \mathbb{R}$. Prove that $ f\left(x \minus{} f\left(x\right)\right) \equal{} 0,\forall x\in \mathbb{R}$[/quote]

1) how could you write "easy" if you dont know the solution (since you posted in unsolved forum)

2) better to indicate the origin in the subject of the problem rather than "easy". For example "First round-greek mathmatical olympiad 2009, problem 2-12 class"

3) better to search a bit before posting : see http://www.mathlinks.ro/Forum/viewtopic.php?t=313557

[quote="pco"][quote="Γιώργος"]The following is the second problem from the first round of the greek competition.

Let $ f: \mathbb{R}\longrightarrow \mathbb{R}$ be a function and it holds that $ f\left(x \minus{} f\left(y\right)\right) \minus{} f\left(y \minus{} f\left(x\right)\right) \equal{} 2f\left(f\left(x\right) \minus{} f\left(y\right)\right), \forall x,y\in \mathbb{R}$. Prove that $ f\left(x \minus{} f\left(x\right)\right) \equal{} 0,\forall x\in \mathbb{R}$[/quote]

1) how could you say "easy" if you dont know the solution (since you posted in unsolved forum)

2) better to indicate the origin of the problem rather than "easy". For example "First round-greek mathmatical olympiad 2009, problem 2-12 class"

3) better to search a bit before posting : see http://www.mathlinks.ro/Forum/viewtopic.php?t=313557[/quote]

I'm terrible sorry for my mistake. Thank you Patrick for the remark. Hope never do that again :(

[quote="Γιώργος"] I'm terrible sorry for my mistake. Thank you Patrick for the remark. Hope never do that again :([/quote]

That's not so terrible  :lol: