A coin is flipped 10 times. What is the probability (as a fraction) that is will land successfully on heads 10 times in a row?

[b]I think it's:[/b]
[hide]$\frac{1}{240}$[/hide]

Try to explain your answers and notify me if mine's wrong.(I get extra credit in math for doing a good job :lol:  :blush: )

Thanks!

there are $2^{10}=1024$ total outcomes. You are choosing 10 to be heads so it is:
$\frac{\binom{10}{10}}{1024}=\boxed{\frac{1}{1024}}$

[hide="or.."]the probability that we get a head is 1/2.  The probability that 10 heads happen (or any particular 10 coin flip sequence happening) is (1/2) * (1/2) * ... * (1/2) * (1/2) = (1/2)^10 = $\boxed{1024}$[/hide]

[b]Thanks.[/b]

I kept trying to find a pattern in listing the possibilities.
If heads was 1, and tails is 0,
For 10, it should be 1,1,1,1,1,1,1,1,1,1.
For 9, it could be 1,1,1,1,1,1,1,1,1,0. There are 10 possibilities with 9 heads and 1 tail.

Well, eventually after adding the numbers together, $1+10+10+9....$ the result gave me a probability of $\frac{1}{240}$. That was about $0.416\%$ which seemed about right though.

The first part of your explanation is right.(the first 1 and 10) after that, the other numbers are : $\binom{10}{3}, \binom{10}{4}, \binom{10}{5}, \binom{10}{6}, \binom{10}{7}, \binom{10}{8}, \binom{10}{9}, \binom{10}{10}$
all of these numbers sum up to 1024. only one works
therefore, the answer is 1/1024


[quote="mdk"]If r_1, r_2, r_3, r_4 are the roots of x^4-4x^2+2=0, what is the value of (1+r_1)(1+r_2)(1+r_3)(1+r_4) ?[/quote]

Right from Math League #6 Contest 3.

[hide]Consider the polynomial

$f(x)=(x-r_{1})(x-r_{2})(x-r_{3})(x-r_{4})=x^{4}-4x^{2}+2$

Now figure out what I am implying[/hide]

[quote="mdk"]If $r_{1}, \ r_{2}, \ r_{3},\ r_{4}$ are the roots of $x^{4}-4x^{2}+2=0$, what is the value of $(1+r_{1})(1+r_{2})(1+r_{3})(1+r_{4})$ ?[/quote]

[hide]
When you expand the expression you get $1+\Pi_{1}+\Pi_{2}+\Pi_{3}+\Pi_{4}$, where $\Pi_{n}$ is the $\text{n}^\text{th}$ symmetric polynomial. We can then just use Vieta's to plug in numbers; the expression is equal to $1+0+(-4)+0+2=\boxed{-1}$.[/hide]

Yep, -1 is the correct answer so it doesn't really matter.

[quote="mdk"]I got -1, but I did it the stupid way. This is a lot easier, and I know Vieta, but the expansion got me stumped. Thanks.[/quote]

No prob.  :)   (Oh, and if you post problems directly from a contest or other source, can you say so in the "Subject Description" under the title? Thanks.  :ninja: )

[hide]

$r_{1}+1, r_{2}+1, r_{3}+1$, and $r_{4}+1$ are the roots of the equation $(x-1)^{4}-4(x-1)^{2}+2=0$ The constant term on this polynomial is the polynomial evaluated at $x=0$

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

Since the constant term is the product of the roots, the answer is $\boxed{-1}$

[/hide]

Nice problem, need to know Vieta to do this.  Can anyone tell me when else you can use Vieta?


algebra
polynomial
logarithms
number theory
prime factorization
number theory proposed

For any $C>0$, show that there exists some $n > 0$ such that $n^{2}+1$ has a prime factor bigger than $Cn$. :)

Very good! That was simple... Try the more general problem: for every nonlinear integer polynomial $P(n)$, there exists (for any $C>0$) some $n>0$ such that $P(n)$ has a prime factor $>Cn$.

[quote="cuopbientcd"]I think i can adopt  $n=2^{2^{m}}$  ,  therefore $n^{2}+1=2^{2^{m+1}}+1$. all prime factor of it has form $2^{m+2}k+1$[/quote]

How does this work, exactly? :? Perhaps I am missing something obvious? We have prime factors of the form $2^{m+2}k+1$. We want prime factors larger than $Cn=C2^{2^{m}}$. $2^{2^{m}}$ is much larger than $2^{m+2}$, but as I said, maybe the idea behind it is subtler and I am just not getting it.

The general problem is equivalent to proving that there are infinitely many primes $p$ for which the polynomial $P$, regarded as a polynomial with coefficients in $\mathbb Z/p$, has at least one root among the smallest $\frac{p}{C}$ residues modulo $p$. My first idea when I first saw the problem (that was a couple of minutes ago :)) was to try to prove that the density of the set of primes with this property is positive, but I am not aware of such a density result. I'll try some more.

P.S.

The same problem was solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=dirichlet&t=28606]here[/url], for $P(x)=x^{4}+1$ and $C=2$. The exact same technique works for $C=2$ and any polynomial which can be written as a polynomial of $x^{2}$. This doesn't help at all for the general problem, of course; it's just a useless (as is most of this message) observation :).

[quote="grobber"][quote="cuopbientcd"]I think i can adopt  $n=2^{2^{m}}$  ,  therefore $n^{2}+1=2^{2^{m+1}}+1$. all prime factor of it has form $2^{m+2}k+1$[/quote]

How does this work, exactly? :? Perhaps I am missing something obvious? [/quote]

Sorry -- you are right  :blush:. We know that $2^{2^{m+1}}+1$ has prime factorization $(2^{m+2}k_{1}+1) \cdots (2^{m+2}k_{r}+1)$, but we can't assume that this number is squarefree, so that we could have $k_{1}< \cdots < k_{r}$... . If we could use "assume" there are arbitrarily large $N$ with $2^{2^{N}}+1$ squarefree, then we could easily construct an argument along these lines (and maybe this holds for all $N$, but it's impossible to prove such things with any current methods...). But... such is life :)

Sorry, I wasn't thinking  :blush:. You are right, this is an invalid argument. So, the question for an *elementary* treatment with $n^{2}+1$ and arbitrarily large $C$ still remains. The general result is by Nagel.

By the way, I had forgotten about the previous discussion :)

[quote="grobber"] The exact same technique works for $C=2$ and any polynomial which can be written as a polynomial of $x^{2}$. [/quote]

Haha :) That's a beautiful 1-line solution in that case  :lol:

By the way, for the $n^{2}+1$ problem, more is known. Hooley has applied bounds on Kloosterman sums to show that for infinitely many $n$, $n^{2}+1$ has a prime factor bigger than $n^{11/10}$. Unfortunately, his methods (not elementary, because the involved bounds on Kloosterman sums aren't quite) don't work for other polynomials...

But the simpler problem with $Cn$ is indeed elementary. Not so much as the nice special case you cite, though :)

I think $Cn$ is mean $C_{n}$ .
But if it is $C.n$ then ?

Ok, I hope I have a completely elementary solution. Let $A$ be the set of the primes of the form $4k+1$. Here's what I did: obviously, we need an upper bound for $\sum_{1\leq i\leq n}{v_{p}(i^{2}+1)}$ (you'll see why). Now, if $n_{i}$ is the number of $j$ with $p^{i}| j^{2}+1$, then the above sum is just the sum of all $n_{i}$ for $i=1,2,...,\log_{p}(n^{2}+1)$. Now, observe that if $p>2$ $p^{i}|j^{2}+1$ and $p^{i}|k^{2}+1$, then $p^{i}$ divides $j-k$ or $j+k$ (this follows from the fact that $p^{i}$ divides $(j-k)(j+k)$ and $p$ does not divide $j,k$). Thus, if $j_{1}$ is the smallest with $p^{i}|j^{2}+1$, then all those $k$ with $1\leq k\leq n$ for which $p^{i}|k^{2}+1$ are either of the form $j_{1}+sp^{i}$ for some $0\leq s\leq\frac{n-j_{1}}{p^{i}}$ or of the from $-j_{1}+tp^{i}$ for some $\frac{1+j_{1}}{p^{i}}\leq t\leq\frac{n+j_{1}}{p^{i}}$. Thus anyway, $n_{i}\leq 3+\frac{2n}{p^{i}}$. Now, if $p=2$, observe that with the same argument if $2^{i}$ divides $j^{2}+1$ and $k^{2}+1$, then $2^{i-2}$ divides $j-k$ or $j+k$. So $\sum_{i=1}^{n}{v_{2}(i^{2}+1)}\leq Cn$ for some absolute constant $C$. Also, $\sum_{1\leq i\leq n}{v_{p}(i^{2}+1)}\leq 3\log_{p}(n^{2}+1)+\frac{2n}{p-1}$. Ok, now let $p(n)$ be the largest prime divisor of at least one of the numbers $1^{2}+1,2^{2}+1,...,n^{2}+1$. Thus, $2\log n!<\log (1^{2}+1)(2^{2}+1)...(n^{2}+1)=\sum_{p\in A,p\leq p(n) or p=2}{\log (p)\cdot (\sum_{1\leq i\leq n}{v_{p}(i^{2}+1)})}$ $<\sum_{p\in A, p\leq p(n)}{\log(p)(3\log_{p}(n^{2}+1)+\frac{2n}{p-1})}+C\log 2\cdot n$. Thus $2\log(n!)<2n\cdot\sum_{p\leq p(n), p\in A}{\frac{\log(p)}{p-1}}+3\pi(n)\log(n^{2}+1)$.  It means that there exists a constant $c$ such that $2\log(n)<2\sum_{p\leq p(n), p\in A}{\frac{\log(p)}{p}}+c$, with absolute constant(because there are absolute constants $k,r$ with $\pi(n)<kn/\log(n)$ and $\log(n!)>n\log n-rn$. Now, using the fact that the sum of $\log p/p$ for primes $p\in A$ and not in $A$ are equivalent (the sum for $p\leq x$), I obtain that $2\log(n)<\log p(n)+o(\log p(n))$. And the problem is that this gives $p(n)>n^{2-\epsilon}$, which is far better than $Cn$, but which surely must be wrong. Unfortunately, I am too tired now to see the mistake(s). Please, can someone point out the stupidity?

[quote="harazi"] Thus, $2\log n!<\log (1^{2}+1)(2^{2}+1)...(n^{2}+1)=\sum_{p\in A,p\leq p(n) or p=2}{\log (p)\cdot (\sum_{1\leq i\leq n}{v_{p}(i^{2}+1)})}$ $<\sum_{p\in A, p\leq p(n)}{\log(p)(3\log_{p}(n^{2}+1)+\frac{2n}{p-1})}+C\log 2\cdot n$. Thus $2\log(n!)<2n\cdot\sum_{p\leq p(n), p\in A}{\frac{\log(p)}{p-1}}+3\pi(n)\log(n^{2}+1)$. [/quote]

But... $\sum_{p \in A, p \leq p(n)}\log(p) = \pi(p(n))/2+o(\pi(p(n)))$, no?... and the estimate you get actually has nothing to do with $\pi(n)$; I think it should be more like $2\log{n}< c_{1}+2\sum_{p \in A, \, p \leq p(n)}\frac{\log{p}}{p}\,+\, c_{2}\frac{p(n) \log{n}}{n\log{p(n)}}$. This gives $2 < \frac{\log{p(n)}}{\log{n}}+c_{3}\frac{p(n)}{n\log{p(n)}}$. This doesn't quite imply $p(n) \gg n^{2-\varepsilon}$ (the latter, to the best of my knowledge, is wide open!), nor anything like $p(n) > n^{c}$ for any $c>1$; but it does show the desired $p(n) > Cn$ for $n \gg_{C}0$. So, very nice approach! :)

Am I correct?  :maybe:

What is your $c>1$ explicitly? I don't know what is the best result to date, but I know of a result of Iwaniec which gives $c = 6/5$.

By the way, your proof easily extends to all polynomials for the $Cn$ problem. Could your strengthening with $c>1$ generalize as easily?? As far as I know, it is not known that for every polynomial $f(n)$ there is a $c_{f}> 1$ such that $f(n)$ has a prime factor $>n^{c_{f}}$ for infinitely many $n$; let alone a universal $c > 1$... I think that proving something like this would be quite a feat. Of course, I may be wrong.  :maybe:

If you like to share your solution for $n^{2}+1$ and $c>1$, I'd be happy to see it :)

I don't understand the confusion here - Is the problems:

a. For all constant $c >0$, there exists infinitinely many positive integers $n$ such that the maximal prime divisor of $P(n) = n^2+1$ is greater than $c*n$.
b. Replacing $P(n)$ by any integral polynomial of degree >1, same thing holds

both true (okay, I know even stronger version of a by replacing $cn$ by $c*n*log \; n$ holds true, but the solution uses overkill theorems), and solvable by elementary methods ? (not resorting to PNT/Dirichlet etc)


$S_k=\{ (u_n) \; non-zero\; real \; sequence \; | \forall n \in N, u_{n+k}=u_n+u_{n+k-1} \}$

Find all $k$ positive integer such that $S_k$ contains a periodical
sequence, find all periodical sequence of $S_k$

There has to be an extra condition around here somewhere, since all $S_k$ contain a periodic sequence, namely the constant sequence $u_n=0,\ \forall n$.

As grobber said the sequence of $S_k$  is not constant $0$ sequence

thx


Two cards are dealt at random from a standard deck of 52 cards.  What is the probability that the first card is a $ \diamondsuit$ and the second card is an Ace?

Assuming replacement, I think your solution is fine.

I think it works even without replacement:
Case 1:
First card is diamond other than ace=3/13
Second is ace=4/51
So 4/221

Case 2:
First card is ace of diamonds=1/52
Second is ace=1/17
So 1/884

Adding 4/221 to 1/884 gives 17/884=1/52.

[quote="bonapup"]Hey I'm a little confused, I found a completely different solution from the one Alcumus had, with the same answer. I did it this way: There is a 1/4 probability that the first card is a diamond. Then there is a 1/13 probability that the next card is an ace. 1/4 x 1/13 = 1/52. did I do it right?
 :huh:
~Slightly confused[/quote]

Me too.

[hide="Sorry for reviving...."]
But idk what is wrong with my solution:

2 cases:

Case 1. 12 choices for a diamond and 4 choices for an ace.
Case 2. 13 choices for a diamond and 3 choices for an ace.

$12*4=48$

$13*3=39$

Total $52*51/2=1326$

$87/1326=$ $\boxed{29/442}$

The answer key says $1/52$, but I don't know what I did wrong.
Help? :help:[/hide]

[hide="Solution"]
[hide="Assuming with replacement"]The first card picked must be a diamond, of which there are 13 out of 52 cards to choose from: $\frac{1}{4}$.

The second card picked must be an ace. There are 4 aces out of 52 cards: $\frac{1}{13}$.

In order for both of these to happen, the probability must be multiplied: $\frac{1}{52}$[/hide]

[hide="Solution without replacement"]I'll take a stab at this,......not guaranteed right answer though.

The first card has $\frac{1}{4}$ probability of being a diamond.

Now we have two cases:
[hide="If the diamond is an ace"]We have 3 more aces to choose from. The probability is $\frac{3}{51}=\frac{1}{17}$.[/hide]
[hide="If the diamond is not an ace"]We have $\frac{4}{51}$. [/hide]

Therefore the probability for no replacement is $\frac{1}{4} \cdot \frac{7}{51}=\frac{7}{204}$.

[/hide][/hide]

[hide]I did it this way: There are 52*51 total choices for the cards, and there are 13 ways to get a diamond, and 4 ways to get an ace. However, you are counting the ace of diamonds twice, so I just subtracted 1. That gave 13*4-1/52*51 = 51/(52*51) which is obviously 1/52. It's right, but is this not a good way to do it?[/hide]

[hide=Easy Sol]Number of diamond cards = 13
Number of ace cards = 4.
Probability of drawing diamond first = 1/4
Probability of drawing ace first = 1/13.
1/13*1/4 = 1/52.[/hide]


This is just an announcement.

The COP 4th will start in 4/29/05 which is about 3 weeks from now.

I can start a week earlier but I'm hoping to build up another short-list where more people can practice their skills on it. :)

Yes.

Please read the About COP thread before asking here! :) It's all there.

This Friday is the date that test will be posted.


factorial
parameterization
probability
expected value
probability and stats

I have some questions about this 2-way analysis of variance

What does balanced mean?

What is the difference between completely randomized design and a blocked design, include which one would block?

What is the model? Explain, what each term represents and what assumptions, if any, are usually made?

What are the hypotheses to be tested?

These are the answers I came up with, does anybody see anything wrong with any of them? Please let me know if there is

What does balanced mean?

[color=red]Means that every cell has the same number of observations[/color]

What is the difference between completely randomized design and a blocked design, include which one would block?

[color=red]A completely randomized design consists of a design where the subjects are assigned completely randomly to the groups. A blocked design, on the other hand, corresponds to a design where the subjects are divided into categorized groups with the main purpose of reducing the variance.[/color]

What is the model? Explain, what each term represents and what assumptions, if any, are usually made?

[color=red]This is a two-way factorial ANOVA. We have two factors A and B, and we have a number of replications for each, which means that the interaction term A*B can also be analyzed. The assumptions made are pretty much the same as for a one way ANOVA: It is assumed that the subjects are assigned randomly and independently to the groups. Also we require normality and homogeneity of variances.[/color]

What are the hypotheses to be tested?

[color=red]That the null hypotheses are correct: [/color]

[color=red]$ H_{0A}: \alpha_1 \equal{} .... \equal{} \alpha_i \equal{} 0$[/color]

[color=red]$ H_{0B}: \beta_1 \equal{} .... \equal{} \beta_j \equal{} 0$[/color]

[color=red]$ H_{0AB}: \gamma_{ij} \equal{} 0$ for all i, and j[/color]

[color=red]where $ \gamma_{ij}$ are the interaction parameters, $ \alpha_i$ are the main effects for factor A and $ \beta_j$ are the main effects for factor B.[/color]

Complete the following table:

[b]Source:[/b]
[b]A[/b]           [b]Degrees of Freedom:[/b] 2  [b]Sum of Squares:[/b] 308.25 [b]Mean of Squares:[/b] [color=red]154.13[/color] [b]F:[/b]  [color=red]8.31[/color] [b]p-value:[/b] [color=red]0.00061[/color]
[b]B[/b]           [b]Degrees of Freedom:[/b] 4 [b]Sum of Squares:[/b] 480.52 [b]Mean of Squares:[/b] [color=red]120.13[/color] [b]F:[/b]  [color=red]6.48[/color] [b]p-value:[/b] [color=red]0.000188[/color]
[b]AB[/b]         [b]Degrees of Freedom:[/b] [color=red]8[/color] [b]Sum of Squares:[/b] 130.1 [b]Mean of Squares:[/b] [color=red]16.26[/color] [b]F:[/b] [color=red]0.88[/color] [b]p-value:[/b] [color=red]0.51[/color]
[b]Error[/b]     [b]Degrees of Freedom:[/b] [color=red]65[/color] [b]Sum of Squares:[/b] [color=red]1205.27[/color] [b]Mean of Squares:[/b] [color=red]18.54[/color]
[b]Total[/b]     [b]Degrees of Freedom:[/b] 79 [b]Sum of Squares:[/b] [color=red]2124.14[/color]

The values in red are the values I came up with

Since this is a balanced design, how many observations are made at each combination of A and B?

[color=red]There are 3x5 = 15 cells and 80 observations, so each cell has 6 replications.[/color]

[color=blue]This is the answer I came up with, but since the number of total observations is 80 (since ijk-1=79), wouldn't the degrees of freedom have to be 89?[/color]

Using the p-values you obtained, state what you would do next.

[color=red]We have to check whether or not the factors A and B (and their interaction) are significant.[/color]

What are your conclusions (expressed for a non-statistician)?

[color=red]The conclusion in that both factors A and B are significant (p < 0.01), but the interaction AB is not (p > 0.1).[/color]

Now assume the p-value for AB is significant. Restate your conclusions.

[color=red]In that case, the significance of each individual factor is disregarded, because now their effects are interrelated.[/color]

[quote="223rem"]
What is the model? Explain, what each term represents and what assumptions, if any, are usually made?

[color=red]This is a two-way factorial ANOVA. We have two factors A and B, and we have a number of replications for each, which means that the interaction term A*B can also be analyzed. The assumptions made are pretty much the same as for a one way ANOVA: It is assumed that the subjects are assigned randomly and independently to the groups. Also we require normality and homogeneity of variances.[/color]
[/quote]

Also, is the model for this 2-way ANOVA as follows:

$ X_{ijk} \equal{} \mu \plus{} A_i \plus{} B_j \plus{} G_{ij} \plus{} \epsilon_{ijk} \newline i \equal{} 1,....,I, j \equal{} 1,....,J, k \equal{} 1,.....,K$

where $ A_i's$, $ B_j's$,  $ G_i's$, and $ \epsilon_{ijk}$ are independent and normally distributed random variables, each with an expected value 0 and variance $ \sigma_A^2$, $ \sigma_B^2$, $ \sigma_G^2$, and $ \sigma^2 respectively$.

[quote="223rem"]
What are the hypotheses to be tested?

That the null hypotheses are correct:

$ H_{0A}: \alpha_1 \equal{} .... \equal{} \alpha_i \equal{} 0$

$ H_{0B}: \beta_1 \equal{} .... \equal{} \beta_j \equal{} 0$

$ H_{0AB}: \gamma_{ij} \equal{} 0$ for all i, and j

where $ \gamma_{ij}$ are the interaction parameters, $ \alpha_i$ are the main effects for factor A and $ \beta_j$ are the main effects for factor B.
[/quote]

On this one, I think I was wrong since it is a completely randomized 2-way ANOVA, so the correct answer would be:

What are the hypotheses to be tested?

We are testing the null hypotheses versus the alternate hypotheses:

$ H_{0A}: \sigma_A^2 \equal{} 0$ versus  $ H_{aA}: \sigma_A^2 > 0$

$ H_{0B}: \sigma_B^2 \equal{} 0$  versus  $ H_{aB}: \sigma_B^2 > 0$

$ H_{0G}: \sigma_G^2 \equal{} 0$  versus  $ H_{aG}: \sigma_G^2 > 0$

where $ \sigma_A^2$, $ \sigma_B^2$, $ \sigma_G^2$ are the variances of factor A, B, and the interaction term A*B respectively. It is customary to test $ H_{0A}$ and $ H_{0B}$ only if $ H_{0AB}$ can not be rejected.

Can anybody corroborate this, it is important to me. Thanks for your help.


Να βρεθούν όλα τα τρίγωνα των οποίων οι πλευρές ικανοποιούν:

$ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\equal{} \frac {9}{1 \plus{} 2abc}$

Κώστα έκανες εξαιρετική δουλειά  :D  ΜΠΡΑΒΟ !

Λοιπόν θα δείξω ότι $ \sum\frac {1}{a^2}\geq\frac {9}{1 \plus{} 2abc}$ για κάθε τρίγωνο
(και το ζητούμενο τρίγωνο θα είναι το ισόπλεευρο με πλευρά 1)

Κάνουμε το δυικό μετασχηματισμό $ a \equal{} x \plus{} y$ κτλ οπότε έχουμε να δείξουμε ότι

$ \sum\frac {1}{(x \plus{} y)^2}\geq\frac {9}{1 \plus{} 2(x \plus{} y)(y \plus{} z)(z \plus{} x)}$

Από την περίφημη Iran 1996 έχουμε $ \sum\frac {1}{(x \plus{} y)^2}\geq\frac {9}{4(xy \plus{} yz \plus{} zx)}$

Επομένως μένει νδο $ 1 \plus{} 2(x \plus{} y)(y \plus{} z)(z \plus{} x)\geq 4(xy \plus{} yz \plus{} zx)$

Είναι επίσης γνωστό ότι

$ 2(x \plus{} y)(y \plus{} z)(z \plus{} x)\geq\frac {16(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)}{9}\geq\frac {16\sqrt {3(xy \plus{} yz \plus{} zx)}(xy \plus{} yz \plus{} zx)}{9}$

Επομένως αρκεί ($ xy \plus{} yz \plus{} zx \equal{} p$)  $ 1 \plus{} \frac {p\sqrt {p}}{9}\geq 4p$  ή ισοδύναμα

$ (\sqrt {p} \minus{} \frac {\sqrt {3}}{2})^2(\sqrt {p} \plus{} \frac {\sqrt {3}}{4})\geq 0$ που ισχύει so a big QED  :D

Σ' άρεσε ε? Πολύ ωραία και η λύση! :)

Ναι η κατασκευή είναι εξαιρετική . Ειδικά η προσθήκη της ανισότητας που η ισότητα ισχύει για όλα ίσα με 1 είναι πολύ καλή κίνηση που δημιουργεί ανομοιογένεια κάνοντας το πρόβλημα εξαιρετικά ενδιαφέρον . Μπράβο και πάλι

Και να φανταστείς δεν την έλυσαν στο inequalities section. Τους έκανα link για εδώ... :)

Η δική μου λύση χρησιμοποιεί την γνωστή ανισότητα $ x\in (0,3]$ και για $ a,b,c\ge0$ έχουμε $ 3\minus{}x\plus{}x(abc)^\frac{2}{x}\plus{}\sum a^2\ge2\sum ab$
και μετά την πασίγνωστη του Iran.Είναι αρκετά όμοια με του silouan.

η δικια μου λυση εχει ως εξης,αφου a,b,c>0
$ LHS >\equal{} \frac{3}{\sqrt[3]{(abc)^2}} >\equal{} \frac{9}{1\plus{}abc\plus{}abc}\equal{} \frac{9}{1\plus{}2abc}$
αφου μονο την ΑΜ-ΓΜ εχω ισοτητα για α=β=γ και αβγ=1 =>α=β=γ=1 :)

ΟΧ!!! Τώτα κατάλαβα γιατί δεν την έλυναν στο inequalities section. Νόμιζαν ότι τους έκανα πλάκα. :(

Athinaios έχεις απόλυτο δίκιο. Λύνεται κι έτσι με χαζό AM-GM. :(  Πώς τυφλώθηκα έτσι και τη χαλάρωσα. Το ωραίο είναι ότι παρέσυρα και silouan και Vaf... :rotfl:


geometry
trigonometry
quadratics
circumcircle
geometry unsolved

Let ABC be atriangle and D be the mid point of side BC. Suppose <DAB=<BCA and <DAC=15. If O is the circumcentre of ADC, prove that triangle AOD is equilateral.
PS: Its easy using trignometry.Can somebody pls give a Geometric proof..

is little bit of trignometry allowed :maybe:

Its simple using sine rule, I guess he wants a proof using "no trigonometry"

[quote="pardesi"]is  allowed :maybe:[/quote]

its easy if we use sine rule.. i was trying to find some construction..
Anyways what is your solution with a "little bit of trignometry"

My method is a bit longish... write sine rule for the two triangles ABD and ADC after we obtain some angles. Keeping x as angle ACBThen we get a quadratic in tan x, which when we solve we get x=30, so the problem is done.

Iam sure there must be a purely Geometric solution to it especially since AB is tangent to the circumcircle of ADC

u can surely reduce the trignomettry content...but would have to use law of sines directly or indirectly :wink:

Δ ACD ≡ Δ ADE
ADXAD = ACXAE &  AD = √2AE
Construct ED’ = AE and perpendicular to AE.
 AD’=AD and Δ ADD’ is Equilateral
Perpendicular Bisectors of AD and AC will intersect at D.
Therefore D’ is the Circum Centre of Δ ACD.
And hence the result.


function Putnam algebra polynomial calculus integration derivative

[color=cyan]This is invented by a friend of mine who is very smart (Putnam Fellow 2 years ago) but not immensely familiar with generating functions.  So it may be easier than he thought.  He thought it was doable, but hard enough to be fun.  I don't really know genfuncs at all, so I can't say.  Anyhow.

Compute the generating function for C(n, 3r).

(Also, of course, you can do other examples first.)[/color]

Are you asking for a generating function of the form :Sigma: (n choose 3r) x^n, :Sigma: (n choose 3r) x^r, or :Sigma: (n choose 3r) x^r*y^n?

The generating function of terms of the form a(k) is the function which has polynomial expansion

f(x) = a(0) + a(1)x + a(2)x^2 + a(3) x^3 + ... + a(k) x^k +...

For example, the generating function

1/(1-x) = 1 + x + x^2 + x^3 + x^4 + ...

is the generating function for a(k) = 1.

Square the above to see that the generating function for a(k) = k+1 is 1/(1-x)^2.

We will cover generating functions and their application to problem solving in our Intermediate Counting class starting at the end of September.

Note that the above series for the function 1/(1-x) only converges on the interval (-1,1).

Right. Often when we work with generating functions, we ignore convergence issues and deal with what we call formal power series. If the series is not convergent, we can't do quite as much with it as we could if it were convergent, but it can still be very useful.

I really need to sign up for that counting class, I know so little...

Sunday is a good day for me.

[color=cyan]Simon:  Define C(n, r) = n(n-1)(n-2)...(n-k+1)/k! for integral k and real n.  Let a_k = C(n, 3k).  Find
f_n(x) = a_0 + a_1(x) + a_2(x^2) + . . .
which will just be polynomial for positive integral n.[/color]

I love this fact:

1/3 (1 + (e^(2i*pi/3))^n + (e^(4i*pi/3))^n) = 1 if n is a multiple of 3 and 0 otherwise.

Thus the generating function for :Sigma:_k (n choose 3k) x^(3k) = 1/3 ((1+x)^n + (1+e^(2i*pi/3)x)^n + (1+e^(4i*pi/3)x^n).

[color=cyan]Ooops . . . there was another condition on the problem:  no using complex numbers.  Oh, well, too late now I guess.[/color]

I'm not really sure if you can do it without in a nice closed form like that without complex numbers. I forget what that theorem on analytic continuation is called, but you have a unique analytic continuation in a region of the complex plane if it is fixed on some curve within that region (in this case the real line). I don't see those powers and getting a closed form, even though the complex numbers will all cancel out.

Hey, can you find me a nice expression for x + sqrt(2)x^2 + sqrt(3)x^3 + 2x^4 + ... ?

Is there a nice way of dealing with recurrences like a_(n+1) = (a_n)^3 + 1?

You really think that you can put all those nasty square roots together and get some nice expression? I just don't see it.

I'll work on the other one, but I don't know any techniques for solving non-linear recursions in general.

I think the expression 1 + :rt2:x:^2: + :rt3:x:^3: + . . . is the definition of a function, sort of . . .
at least, mathematica turns it into a function whose definition is a bit beyond me.  Go to mathworld.wolfram.com and look up polylogarithmic functions -- it'll tell you all about it, I expect.

[color=cyan]Okay, I got it.  It's the polylogarithmic function
Li_n(x) = 1 + 2^(-n)x^2 + 3^(-n)x^3 + . . .
or in other words, it defines your series as itself.  To find if it has any worthwhile properties, go see mathworld.[/color]

Well, you know, it's not fair; we can get a polynomial P(n) by using P(xD_x)1, but there's no nice expression for even the simplest fractional exponent? Differentiation is just a transformation of a polynomial, there ought to be something that is analogous to its square root.

It's not fair.

You have to be careful about where these series converge sometimes. Otherwise you can get some things that don't make sense. Even if you're using them in just a formal manner, ideally you would like it to converge in some interval of the real line (or region of the complex plane).

In the case of this series with the square roots, it doesn't converge for any real (or complex) values other than zero. So you're not going to get any function that equals this series.

The polylogrithmic function only converges on a real interval when n>=0. But to equal the series with the square roots, n = -1/2.

Are you sure about that gauss202? I think the series with the square roots converges is |x|<1. As for the differentiation transformation, I was thinking maybe we can take half a derivative, but I don't know enough about them. I played around with them a bit 2 years ago, but I didn't get too far.

ah. you're right, it does converge for |x| < 1. In that case, I think it should match this polylogrithmic function! But the convergence *somewhere* is a big key. Because then you can use analytic continuation to extend it elsewhere. And even if you use it as a formal power series, that gives it a lot more grounding.

Here is a brief but not very good way of thinking of analytic continuation:

Suppose a function is defined in a certain way in a certain region of the complex plane. An analytic continuation of that function is defined in that region of the complex plane as well as some more, and it gives the same values in the region where they're both defined.

Example: 1+x+x^2+x^3+... is defined for |x|<1. Its analytic continuation, analytic everywhere except at x=1, is 1/(1-x).

uh, can you explain that more in detail,

or direct me to some references?

im kinda slow, and terrible at functions, so you might need some patience....

Basically, a function is analytic if it has continuous derivatives of every order. So if you can differentiate it an arbitrary number of times and those derivatives are all continuous then the function is analytic. A theorem of complex analysis is that analytic functions in the complex plane are very rigid. Once you know the values of an analytic function in some region then it uniquely defines the function everywhere in the complex plane. In other words, two different analytic functions cannot have the same values on a region in the complex plane without being equal.

So a power series definies an analytic function is whatever region it absolutely converges. But because of this theorem about analytic continuation, even in the region where it does not converge it has a unique extension to an analytic function which it is equal to in the region that it does converge. So like Zeta said, the series f(x) = 1 + x + x^2 + .... absolutely converges only in the region of the complex plane |x| < 1. But it is equal to the function g(x) = 1/(1-x) there. Therefore, you can say the function f has an analytic continuation to the function g on the whole complex plane minus the point 1.

But the key is that you need some region innitially for it to converge absolutely in order to analytically continue it to some larger region. And not all functions can be analytically continued to larger regions. It depends on where it's points of discontinuity (or poles) are. For example, the function :sigma: {n=0, inf}  z^(n!) converges in the region |z| < 1. But it cannot be analytically continued to any function outside of that region. Can you see why?

This idea also leads naturally to the realm of Riemann Surfaces, which we talked about briefly in another thread somewhere on this site.

Yea, thanks for the explanantion gaus, and Joel explained it to me, 2

But i think analytic continuation is a bit.....advanced for little ol' me.

Haw, apparently there exists something called "fractional integro-differentiation." I postulated its existence all by myself, I figure I'm a regular Karl Friedrich Gauss, or something.

Yes, that's right. I was very disappointed after I came up with it to find out that other people already had a lot of theory on the subject, and I could spend hundreds of dollars on textbooks on fractional calculus. What does a regular Gauss mean? Is that in contrast to an irregular one?

The funny thing is that the fractional calculus doesn't seem to help us here. In order to get 0^p + 1^p*x+2^p*x^2 + 3^p*x^3 + ... we apply (xD_x)^p to 1/(1-x). The iteration of that transformation, which we represent as ^p, is not distributive, which means that we may not necessarily be able to simply multiply the semiderivative by sqrt(x) in order to get the series we desire; in fact, we can't. What a drag!


[hide]
First set up an equation.
The recipricol of (1-x) is 1/(1-x). That value is subtracted from one, and it equals 1/(1-x). The equation is:

$1-\frac{1}{1-x}=\frac{1}{1-x}$

Multiply everthing by $1-x$

$1-x -1 = 1$

$1-x = 2$

$-x=1$

$\boxed{x=-1}$[/hide]

[quote="4everwise"]Albatross wants you to find the value of x if 1 minus the reciprocal of (1-x) equals the reciprocal of (1-x).[/quote]

[hide]$1-\frac{1}{{1-x}}=\frac{1}{{1-x}}$
$1=\frac{2}{{1-x}}$
${1-x}=2$
$x={-1}$[/hide]

Use the hide thing when you post a solution, DrunkenMath.

[quote="xdl615x"]Guys there are three of the same solutions. Please try and avoid that. Only post if you have a question, an extension to the question or a diffrent way to solve it.[/quote]
I disagree.

If you have not read the previous solutions, please feel free to post your own original solutions to challenge problems.  That's why we encourage the use of the "hide" tags.  If your original solution happens to almost duplicate a previous solution, that's OK.  [But please don't read another person's solution and then copy or paraphrase it, unless you have something worthwhile to add.]

[quote="4everwise"]Albatross wants you to find the value of x if 1 minus the reciprocal of (1-x) equals the reciprocal of (1-x).[/quote]

[hide]
$1-\frac{1}{1-x}=\frac{1}{1-x}\\1=\frac{2}{1-x}\\1-x=2\\x=-1$

DONE

[/hide]


Five paychecks and envelopes are addressed to five different people. If the paychecks are randomly inserted into the envelopes, what is the probability that [b](a)[/b] exactly one is inserted in the correct envelope and [b](b)[/b] at least one is inserted in the correct envelope?

[hide="b"] Count the complement, which is $!5 = 44$.  The probability is then $\frac{120-44}{120}= \boxed{ \frac{19}{30}}$.

For more information, consult

http://mathworld.wolfram.com/Derangement.html[/hide]

[quote="anyone"][hide]A. There are 5!=120 possible arrangements of the paychecks. The probability that exactly one is correct would be all combination w/ one right and the rest in any order. Since there are four that can be in any of the other four envelopes there are 4!=24 possible combinations of this. So the answer would be 24/120 = 1/5.

Can someone confirm that this is correct?

Don't know how to do B, sorry[/hide][/quote]
Isn't that the solution to B? When you arrange the other 4, there is a possibility that some will be in the correct places.

[quote="anyone"][hide]A. There are 5!=120 possible arrangements of the paychecks. The probability that exactly one is correct would be all combination w/ one right and the rest in any order. Since there are four that can be in any of the other four envelopes there are 4!=24 possible combinations of this. So the answer would be 24/120 = 1/5.

Can someone confirm that this is correct?

Don't know how to do B, sorry[/hide][/quote]

i don't think this is correct. this acounts for at least one correct.
for exactly one correct, all the others must be in the wrong envelopes. the 4! part accounts for all possibilities right or wrong.

[hide="a"] Applying the same principle as my solution to [b]b[/b], this is $\frac{5 \cdot !4}{120}= \frac{9}{60}= \boxed{ \frac{3}{20}}$. [/hide]
anyone's solution is not correct for either part, since it overcounts arrangements where more than one letter is inserted in the correct envelope.


Let $ p(x,y) \in \mathbb{Q}[x,y]$ and ${ p(\pi, {\pi}^2})=0$. How to show, that $ \forall t\in \mathbb{R}$  $ p(t,t^2)=0$?

Let $ f(x) \equal{} p(x,x^2) \in \mathbb{Q}[x]$. If $ f$ is non-zero, $ f(\pi) \equal{} 0$ implies that $ \pi$ is algebraic (by definition), a contradiction (because $ \pi$ is transcendental). Thus $ p(x,x^2) \equiv 0$.

$ p(x,y)\equal{}\sum_{i,j}C_{i,j}x^iy^j$ , $ p(\pi,\pi^2)\equal{}\sum_{i,j}C_{i,j}\pi ^i\pi^{2^j}\equal{}\sum_k \pi ^k (\sum_{i\plus{}2j\equal{}k}C_{i,j})$
Because of $ C_{i,j}$ are rational and $ \pi$ is transcendent we have to have $ \forall k, \sum_{i\plus{}2j\equal{}k}C_{i,j} \equal{}0$, therefore $ \forall t, p(t,t^2)\equal{}0$
More generally, the statement holds if the $ C_{i,j}$ coefficients are algebric numbers .
 :cool:

Thanks a lot. Can you help me with this one?
Suppose $ p(x,y)$ is irreducible polynomial with algebraic coefficients, both $ x$ and $ y$ occur in $ p$. Define the curve $ C_n$ as the set of pairs $ (x,y)$ such that $ p(nx,y^n)\equal{}0$. Let $ v$ be a transcendental number, such that $ (v,e^v)\in C_n$ or $ p(nv,(e^v)^n)\equal{}0$. Then is it true that for every $ q(x,y)$ - polynomial with algebraic coefficients, such that $ q(v,e^v)\equal{}0$ holds $ p(nx,y^n)|q(x,y)$?


Determine the number of subsets $S$ of ${1, 2, 3, . . . , 10}$ with the following property:
there exist integers $a < b < c$ with a $a\in S$, $b\notin S$, $c\in S$.

[hide]This is all subsets of length 3 or more that are not purely sets of consecutive integers.

Sets of length 0 through 2:
(10 0) + (10 1) + (10 2) = 1+10+45 = 56

Sets of length k that are sets of consecutive integers:
(10-k+1)

Total subsets sets of length 3 or more that are sets of consecutive integers: 8+7+6+5+4+3+2+1=36

Total subsets of S = 2^10 = 1024

Number of subsets with this property = 1024-36-56=$\boxed{942}$[/hide]


LaTeX
Euler
number theory unsolved
number theory

Suppose that the positive integers $ a$ and $ b$ satisfy the equation $ a^b\minus{}b^a\equal{}1008$ Prove that $ a$ and $ b$ are congruent modulo 1008.

$ \text{\LaTeX}$ed

Suppose that the positive integers $ a$ and $ b$ satisfy the equation $ a^b-b^a=1008$

Prove that $ a$ and $ b$ are congruent modulo $ 1008$

It had unique solution $ a\equal{}1009,b\equal{}1$.

obviously a and b are of the same parity.
Let a and b be even.Let a>b
IF b is 2 there is no solution, so assume and b to be greater than 2.
Then $ a^b/2 \plus{} b^a/2$ * $ a^b/2 \minus{} b^a/2$ = 1008
But we observe if a and b are even then both  $ a^b/2 \plus{} b^a/2$  and $ a^b/2 \minus{} b^a/2$ are divisible by 4.
Let $ a^b/2 \plus{} b^a/2$ = 4l
and $ a^b/2 \minus{} b^a/2$ = 4m
Then we get  $ 2a^b/2$ is 4(l+m)
So $ a^b/2$ = 2(l+m)
But l and m can be only one of 1,7,9,63
Checking we get no solutions
IF a and b are odd
then $ a^b$ is congruent to $ b^a$ mod 1008
So if $ a^b$ is congruent to $ b^a$ mod 1008 then $ a$ is congruent to $ b$ mod 16,7,9
( i checked all other possibilities by brute force  :rotfl: )
So $ a$ is congruent to $ b$ mod 1008

befuddlers u can try for a more elegant soln.

Woohooo! I was the only one who solved it at the contest  :lol:

Easily $ a \equiv b \bmod 2$. If $ 2 \mid a$ then $ \bmod 32$ we know that $ a,b<5$, and so no solutions. So $ a \equiv b \equiv 1 \bmod 2$. Then $ a^b \equiv b^a \bmod 4$ easily follows, since $ \phi(4) \equal{} 2$, usw. so $ a \equiv b \bmod 16$.
We easily obtain $ a^b \equiv b^a \bmod 3$ since $ a \equiv b \bmod \phi(3) \equal{} 2$. And so $ a^b \equiv b^a \bmod 9$ since $ a \equiv b \bmod \phi(9)\equal{}6$. Since $ \phi(7) \equal{} 6$, we get in the same way $ a^b \equiv b^a \bmod 7$. So $ a^b \equiv b^a \bmod 1008$. QED.

[quote="Rust"]It had unique solution $ a \equal{} 1009,b \equal{} 1$.[/quote]
I think that it is harder to prove than the original statement...

[quote="feliz"]I didn´t understand anything, Mathias... What is the relation between $ a \equiv b mod \phi(n)$ and $ a^b \equiv b^a mod n$? And what do you mean by "$ a^b \equiv b^a mod 1008$. Q. E. D."? (I don´t know how to copy what you wrote) $ a^b \equiv b^a mod 1008$ is obvious![/quote]
Haha.. Sorry of course it is $ a \equiv b \bmod 1008$. I made that mistake a couple of times, just ignore that :)

If $ (a,n) \equal{} 1$, then $ a^{\phi(n)} \equiv 1 \bmod n$. Try searching Eulers phi-function. There's probably a lot of people who can give you a lot better explanation than me ^^

[quote="Mathias_DK"][quote="Rust"]It had unique solution $ a \equal{} 1009,b \equal{} 1$.[/quote]
I think that it is harder to prove than the original statement...[/quote]
Consider $ b\equal{}1$ and get solution $ a\equal{}1009$. Consider $ b\equal{}2$ $ a^2\minus{}2^a\equal{}1008$ had not solution.
Consider $ a\equal{}1$ - had not solution. Consider $ a\equal{}2$ $ 2^b\minus{}b^2\equal{}1008\to 4|b,b\ge 10$, but $ 2^{12}\minus{}12^2>1008$, therefore $ a\equal{}2$ give not solution.
If $ a\ge 3,b\ge 3$, then $ a<b$, because $ a^{1/a}>b^{1/b}$.
Consider $ a\equal{}3$ $ 3^b\minus{}b^3\equal{}1008\to 3|b,a\equal{}3<b<9$, but $ b\equal{}6$ is not solution.
Consider $ a\equal{}4$ $ 4^b\minus{}b^4\equal{}1008\to 2|b, a\equal{}4<b<6$ give not solution.
If $ a\ge 4$, then $ a<b<a\plus{}2$. Therefore may be only $ b\equal{}a\plus{}1$.
$ a^{a\plus{}1}\minus{}(a\plus{}1)^a\equal{}a^a(a\minus{}(1\plus{}1/a)^a)>1008$ if $ a\ge 5$.

[quote="Rust"][quote="Mathias_DK"][quote="Rust"]It had unique solution $ a \equal{} 1009,b \equal{} 1$.[/quote]
I think that it is harder to prove than the original statement...[/quote]
Consider $ b \equal{} 1$ and get solution $ a \equal{} 1009$. Consider $ b \equal{} 2$ $ a^2 \minus{} 2^a \equal{} 1008$ had not solution.
Consider $ a \equal{} 1$ - had not solution. Consider $ a \equal{} 2$ $ 2^b \minus{} b^2 \equal{} 1008\to 4|b,b\ge 10$, but $ 2^{12} \minus{} 12^2 > 1008$, therefore $ a \equal{} 2$ give not solution.
If $ a\ge 3,b\ge 3$, then $ a < b$, because $ a^{1/a} > b^{1/b}$.
Consider $ a \equal{} 3$ $ 3^b \minus{} b^3 \equal{} 1008\to 3|b,a \equal{} 3 < b < 9$, but $ b \equal{} 6$ is not solution.
Consider $ a \equal{} 4$ $ 4^b \minus{} b^4 \equal{} 1008\to 2|b, a \equal{} 4 < b < 6$ give not solution.
If $ a\ge 4$, then $ a < b < a \plus{} 2$. Therefore may be only $ b \equal{} a \plus{} 1$.
$ a^{a \plus{} 1} \minus{} (a \plus{} 1)^a \equal{} a^a(a \minus{} (1 \plus{} 1/a)^a) > 1008$ if $ a\ge 5$.[/quote]
Thanks for your solution on this one :)

[quote="erdugan"]Suppose that the positive integers $ a$ and $ b$ satisfy the equation $ a^b \minus{} b^a \equal{} 1008$ Prove that $ a$ and $ b$ are congruent modulo 1008.[/quote]
Ok. Felix has requested me to clear out my solution, so here it goes:
$ a \equiv b \bmod 2$ is obvious. But $ 2 \nmid a,b$. If $ 2 \mid a,b$ then $ a,b \le 5$ (because of $ \bmod 32$) which gives no solution. Then $ a^b \equiv a \bmod 8$, and $ b^a \equiv b \bmod 8$, so $ a \equiv b \bmod 8$. (Notice $ a^4 \equiv 1 \bmod 16$ if $ (a,2) \equal{} 1$) Then $ a^b \equiv a^a \bmod 16$ so $ a^a \equiv b^a \bmod 16$. If $ a \equiv 1 \bmod 4$ then $ a \equiv b \bmod 16$. If $ a \equiv 3 \bmod 4$ then $ a^3 \equiv b^3 \bmod 16$ $ \Rightarrow$ $ a \equiv b \bmod 16$. (Multiply with $ ab$ and use that $ a^4 \equiv b^4 \equiv 1 \bmod 16$)

Then $ a^b \equiv a \bmod 3$ and $ b^a \equiv b \bmod 3$, because $ a$ and $ b$ are odd. So $ a \equiv b \bmod 3$.
We know $ \phi(9) \equal{} 6$. If $ a \equiv b \equiv 1 \bmod 6$ then $ a^b \equiv a \bmod 9$ and $ b^a \equiv b \bmod 9$ so $ a \equiv b \bmod 9$. If $ a \equiv b \equiv 5 \bmod 6$. Then $ a^b \equiv a^5 \bmod 9$ and $ b^a \equiv b^5 \bmod 9$. So $ a^5 \equiv b^5 \bmod 9$ $ \Rightarrow$ $ a \equiv b \bmod 9$. (Multiply with $ ab$, and notice $ a^6 \equiv b^6 \equiv 1 \bmod 9$).
Since $ \phi(7) \equal{} 6$, we get in the same way that $ a \equiv b \bmod 7$. So $ a \equiv b \bmod 16 \cdot 7 \cdot 9 \equal{} 1008$.


Trivia apart, I think its because it would become a repeat of Davisson and Germer Experiment, which was in many aspects similar to Rutherford but only thing was that they used Ni and e- instead of Alpha particles.

I may be sounding stupid , but
1) do electrons produce scintillations ? DG experiment uses galvanometer to detect current produced by electrons.
2) with their relatively smaller mass, wont they get attracted or repelled much more ?

i don't think so they shud--but excited ones can


algebra
function
domain
real analysis
real analysis unsolved

[quote="J.Y.Choi"]Prove that there exists a bijection that domain $ [0,1]$ and range into $ [0,1]\times [0,1]$.[/quote]

You mean "onto". Hint: $ a_1 a_2 ... \rightarrow a_1 a_3 ..., a_2 a_4 ...$

Here is my approach, please read and point out the error(if there is).

Let's think about a function $ f$ that domain is $ [0,1]$ and range onto $ [0,1]\times [0,1]$ such that $ 0,a_1 a_2 a_3 a_4\cdots \mapsto (0.a_1 a_3 a_5\cdots,0.a_2 a_4 a_6\cdots)$. $ 0.a_1 a_2 a_3 a_4\cdots$ is decimal expression of all elements in the cell $ [0,1]$(so, $ a_i\equal{}0,1,2,3\cdots 9$).
 To prove that $ f$ is a bijection, we have to show that $ (0.a_1 a_3 a_5\cdots,0.a_2 a_4 a_6\cdots)$ can represent all the elements of $ [0,1]\times [0,1]$. To express all the elements in $ [0,1]$, we have to think about all the ways of $ a_i\equal{}0,1,2,3,\cdots 9$. Suppose that $ a_1\equal{}0$, then $ a_3$ can take all numbers $ 0$ to $ 9$, and in each case, $ a_5$ can take all numbers of $ 0$ to $ 9$, and so on. So, $ 0.a_1 a_3 a_5\cdots$ can express all the elements in $ [0,1]$. And we can apply just the same in $ 0.a_2 a_4 a_6\cdots$, so each $ 0.a_1 a_2 a_3 a_4\cdots\in [0,1]$, $ (0.a_1 a_3 a_5\cdots, 0.a_2 a_4 a_6\cdots)$ can represent all the elements of $ [0,1]\times [0,1]$. So, $ f$ is a bijection.

I would say your solution is not precise enough.

You want to show it is surjective and injective. You started with showing it is surjective. Take $ (y,z) \in [0,1] \times [0,1]$ and write them using their (are these unique?) decimal expressions $ y \equal{} 0.b_1 b_2 b_3 \dots$ and $ z \equal{} 0.c_1 c_2 c_3\dots$.
Now you want to define a number $ x \equal{} 0.b_1c_1b_2c_2\dots$, show it is in $ [0,1]$, and show its image is indeed $ (y,z)$.

Showing the map is injective forces you to do deal with a certain problem. This comes from the fact that $ 0.1\overline{0 0 0} \equal{} 0.0\overline{9 9 9}$.

You have not shown your map is injective, and your argument that it is surjective is vague at best. What does this mean: "$ a_5$ can take all numbers of $ 0$ to $ 9$''?
First take an element in the supposed image, and then show you can find (/construct) an element in the domain such that it gets mapped to said element in the image.

What if the problem was the reverse - but you want to show that there is an injection (not bijection) from $ \mathbb{R}^2$ to $ \mathbb{R}$?

I think the spirit of J.Y.Choi's argument is fine, even if the details are not entirely polished.  The issue of non-unique decimal representation is easily dispatched by additionally requiring that $ \liminf _{n\to \infty}a_n(x) < 9$ for all $ x$; with this additional requirement the mapping of real numbers to their decimal representations is bijective.

[quote="aidan"]What if the problem was the reverse...[/quote]Once a bijection $ \phi$ is found on $ [0,1]\to [0,1]^2$ as, say, in Choi's solution, then $ \phi^{ \minus{} 1}$ is a bijection on $ [0,1]^2\to [0,1]$, and then $ \psi^{ \minus{} 1}\left(\phi^{ \minus{} 1}(\psi (x),\psi (y))\right)$ is a bijection (and hence also an injection) on $ \mathbb{R}^2 \to \mathbb{R}$ where $ \psi$ is any bijection $ \mathbb{R} \to [0,1]$.

Then I've got an injection from $ \mathbb{R}^2$ to $ \mathbb{R}$ ^^

J.Y. Choi/Tyl has actually provided an injection from $ [0, 1] \times [0, 1]$ to $ [0, 1]$ - to make it injective we can stipulate that whenever a real number has 2 decimal representations, we take the representation that ends in all $ 9$'s.

To make it surjective as well is a little harder though o.0


geometry
pigeonhole principle
vector
linear algebra
matrix
quadratics
analytic geometry

[b]Edit:[/b] $a,b,c$ are reals and $D=ac-b^{2}>0$. If $f(x,y)=ax^{2}+2bxy+cy^{2}$, prove there exists two integers $u, v$ and $u^{2}+v^{2}>0$ satisfy $f(u, v)\leq \sqrt{\frac{4D}{3}}$

Very strange because it's trivial that $\forall \varepsilon >0 \ \ \exists u \ne 0 \ \ \exists v \ne 0 \ \ (\ |f(u,v)| < \varepsilon )$

Any typo may be?

Try to use Minkowski's theorem. Take it as a hint, I'm not sure it works, but it has big chances...

When I tried Minkowski's theorem (on the area of a centrally-symmetric convex body), I got a bound of $\frac{4}{\pi}\sqrt{D}$, which is a little weaker than what the problem asks for.

Is there any elementary proof (just using the theorem which high school students know)? I don't think Minkowski's theorem is elementary. :idea:

Using Dirichlet's theorem on approximating an arbitrary number by a rational number (which is elementary, just the pigeonhole principle), I got a bound of about $2 \sqrt{D}$, which is even weaker.

Unfortunately, I forgot that for small values of $n$, Minkowski bound is weaker then Hermite's one. Ok, here is the approach for this problem: of course, we may assume that $a,c>0$. Take $v$ to be a nonzero vector which realizes the minimum (call this minimum $m>0$) of $^{t}vAv$ on the set of lattice nonzero vectors (here $A$ is the matrix associated to the quadratic form). Obviously, its coordinates are relatively prime, so we can find an matrix $U\in GL_{2}(Z)$ having $z$ as first column. Using this matrix, we can easily see that $A$ is actually congruent to a symmetric matrix $B=^{t}U AU$ with $b_{11}=m$. Now, by considering the closest integer to $-b_{12}/b_{11}$, we can find an integer $k$ with $|kb_{11}+b_{12}|\leq b_{11}/2$. Take $X$ to be the matrix with lines $(1,0)$ and $(k,1)$. A small computation shows that $X$ has the form $(m,t)$ (first line) and $(t,w)$ with $|t|\leq m/2$ and $w\geq m$. Clearly, $X$ has the same determinant as $A$ (in absolute value), which is $D$. But $D=mw-t^{2}\geq 3/4 m^{2}$, from where the conclusion follows. Of course, one can make this elementary by replacing the above words with explicit computations.


probability
conditional probability
combinatorics solved
combinatorics

You and your (n-1) friends (n > 1) are on a boat that is about to sink.
To lessen the weight of the boat, one of you must jump off. So you all pick a straw from a randomly assorted set of n straws. The one with the short end will be sacrificed5.

Determine when you should draw a straw (as in first, second) to minimize chances of death.

I don't understand...the chance of dying seems to be the same all the time.  Number the straws as 1, 2, ..., n, where straw no. 1 is the short straw.  What each person on the boat gets (assuming you draw straws until the end) is just a permutation of (1, 2, ..., n), and obviously the probability of each permutation is equal, so the probability that you get straw no. 1 is the same no matter which round you draw on...

Or have I misunderstood the question totally?

You're right ^^ I found it interesting that it doesn't matter what order you pick a straw.

Oh... That's it? It seemed to me that the probability is the same no matter when you pick the straw, but I thought I was doing something wrong... :D

An interesting thought related to this problem: The result is so counter-intuitive that it leads to some strange incidents! :D

When asked to draw straws, most people elect to draw as early as possible so that "the probability that I get the short straw is lowest" (as they think), but during balloting among many parents for only a few places to see which children can enter a popular school, most parents want to draw as early as possible "so that my kid will have a higher chance of getting into the school".

Obviously both situations are the same, but most people don't realise it. :)

Here's another non-intuitive problem related to probabilities. You might be familiar with this, since it's pretty famous. It's known as "The Monty Hall problem".

In a game show ("Monty Hall", thus the name of the problem) you have to choose one of 3 gates. A car is hidden behind one of the 3 gates and goates are hidden behind the other 2. You choose a gate. The hoast of the show opens one of the gates with a goat behind them (this gate is different from the one you've chosen) and then asks you if you want to change your choice. Should you or should you not change your choice if you want to have as high a chance as possible to win the car?

Again, it does not matter.

The contestant can either win 1) without changing, or 2) with changing.

The probability of 1) is 1/3.

The probability of 2) is 2/3 * 1/2 = 1/3, since the contestant must have chosen a wrong door on his first try (probability of 2/3) in order for his changed vote to be correct (probability of 1/2).

Either way, the odds are not favorable for the contestant. =)

How did you use the fact that the hoast of the show opens a wrong door after you make your choice? That's why I said it was non-intuitive: the answer is that you should change your mind  :D Think about it.

In fact, let me show you exactly what I mean.

Let's assume you initially choose door 3. The hoast then opens a door which has a goat behind it and offers you the chance to change your initial choice. Let's see what happens if (a) you stick with door 3 or 9b) you change your choice:

(a)
Door 1   Door 2   Door 3   Outcome
car        goat       goat       lose
goat      car         goat       lose
goat      goat       car         win

so you have a 1/3 chance of winning

(b)
Door 1   Door 2   Door 3   Outcome
car        goat       goat       win
goat      car         goat       win
goat      goat       car         lose

so you have a 2/3 chance of winning

The conclusion is that you should change your choice.

The American TV game show where this could be seen was actually called "Let's Make a Deal". "Monty Hall" was the name of the challenge.

Ah I see. My line of reasoning should have stopped at 2/3, since if you assume that you have picked a goat on your first try, the chances of picking the car is 1, and therefore the total chance is 2/3, not 1/3.

I didn't see it either. Intuition tells you "it's the same; it doesn't matter what you do", but we have proof of the fact that intuition is wrong :D

I wonder how many contestants DID change their gates :)

Yeah, there was a game show in Germany too. But it was a slightly modified version. There was no goat but a cuddly toy called zonk. Sometimes there were two zonks as the show master (sm) did not open any door. Prizes ranged from kitchen tools to cars. The sm always tried to confuse you giving you money when you change your decision. And then he says you will get more money if you change back or the third door...

You also can solve the problem in term of elementary probability theory. It is a straight forward application of the following three facts:

Conditional Probability: http://mathworld.wolfram.com/ConditionalProbability.html
Bayes' Theorem: http://mathworld.wolfram.com/BayesTheorem.html
Total Probability Theorem: http://mathworld.wolfram.com/TotalProbabilityTheorem.html

Let T_i be the event that the car is behind door i and let S_i be the event that the sm opens door i. The probability P tells us what the probabilty that behind the door is a car. So we have (1) P(T_i) = 1/3 for i = 1,2,3. Without loss of generality assume that the guest chooses door 3.

case 1: sm opens door 2

(2)  P(T_1|S_2) denotes the probability of T_1 under the condition S_2.
(3)  P(T_3|S_2)

Note that: (2) and (3) adds up to 1.

case 2: sm opens door 1

(4)  P(T_2|S_1)
(5)  P(T_3|S_1)

Now we will study (2) of case 1:

Obviously we have P(S_2|T_1) = 1 (6) due to the case we study.

The Total Probability Theorem for P(S_2) gives:

P(S_2) = P(S_2|T_1)*P(T_1) + P(S_2|T_2)*P(T_2) + P(S_2|T_3)*P(T_3)
= 1 * 1/3 + 0 * 1/3 + 1/2*1/3 = 1/2 (7)

Plugging (1), (6) and (7) in Bayes' Theorem we get:

P(T_1|S_2) = (P(S_2|T_1)*P(T_1))/P(S_2) = (1/3*1)/(1/2) = 2/3.

Hence we get for (3) P(T_3|S_2) = 1/3.

That means we should change the door. Case two can be worked out analogously.

There is also another problem: the prisoner problem. Three prisoners A, B and C are sentenced to death but due to the courtesy of the president one of them will be set free (this one will receive amnesty). A regular ticket will decide who will be set free. But the prisoners won't receive the message who will be set free not earlier than the day of execution. Of course the prisoners are curious. Hence A (similar for B,c) says to the
guard: "One of my fellow prisoners will be killed in any case. So it won't change my situation if you tell me the name of a fellow prisoner that will be killed." The guard thinks about the question and answers finally: "B will be killed".

Let G_i with i = A,B,C be the prisoner to be set free. And W_i with i = A,B,C denote the person to be killed which is mentioned by the guard.

P(G_A) = P(G_B) = P(G_C) = 1/3. (1)

case 1: guard says B to the prisoner A:

P(G_A|W_B) (2)

Due to the case being studied we have P(W_B|G_A) = 1/2. (3)

The Total Probability Theorem for P(W_B) gives:

P(W_B) = P(W_B|G_A)*P(G_A) + P(W_B|G_B)*P(G_B) + P(W_B|G_C)*(G_C) = 1/2 * 1/3 + 1/3 * 0 + 1 * 1/3 = 1/2. (4)

Plugging (1), (3) and (4) in Bayes' Theorem we get (2):

P(G_A|W_B) = (P(W_B|G_A)*P(G_C))/P(W_B) = (1/2 * 1/3)/(1/2) = 1/3 (5)

case 2: guard says C to the prisoner A

P(G_A|W_C), this case is similar to case 1

Due to (5) it is not worth asking the guard whom of the other two guys will be killed.

Monty Hall Problem = MHP and Prisoner Problem = PP

Select the door with the car at MHP is related to be the prisoner to be set free at PP. To choose a different door at MHP would mean that that the prisoners might be able to turn into another person, e.g. A and C will be substituted (A turns C and vice versa). But obviously people cannot be substituted thus the prisoners cannot increase their knowledge whether they will be set free. Am I right ?

some sources:

http://mathworld.wolfram.com/MontyHallProblem.html
http://mathworld.wolfram.com/PrisonersDilemma.html
http://www.gavagai.de/themen/HHPT09.htm
http://barryispuzzled.com/zmonty.htm
http://www.wiskit.com/marilyn.gameshow.html

Here's a page with lots of interesting probability problems:

[url]http://rec-puzzles.org/probability.html[/url]


Alcumus AMC AIME geometry MATHCOUNTS probability Support

Don't you think it would be better for a user to choose his/her own difficulties? What annoys me is that the hardest problem I've seen so far is something around easy AIME level, so wouldn't it be better if there are some really challenging problems (around pre-olympiad level) for people that have bypassed the usual difficulty of Alcumum problems?

Also, wouldn't it be better to have another Alcumus on, say, Geometry (and other subjects, and again, different difficulties for different users)?

(BTW, what does "Alcumus" mean exactly?)

Alcumus is the father of Mentor.

http://en.wikipedia.org/wiki/Alcumus

The AoPS people are going to add geometry, etc., later.

[quote="mathwizarddude"]Don't you think it would be better for a user to choose his/her own difficulties? What annoys me is that the hardest problem I've seen so far is something around easy AIME level, so wouldn't it be better if there are some really challenging problems (around pre-olympiad level) for people that have bypassed the usual difficulty of Alcumum problems?[/quote]

You are assigned problems based on how you did on previous problems; it gives you problems that are suitable for your levels.

Note: Please make sure to read if your question has already been suggested or posted before you make another thread. These questions were mentioned quite a few times.

[quote="r15s11z55y89w21"]You are assigned problems based on how you did on previous problems; it gives you problems that are suitable for your levels.

[/quote]
seriously? if it's suitable for my level then why is it still mathscount level???

BTW when are they going to add geometry?

[quote="mathwizarddude"]seriously? if it's suitable for my level then why is it still mathscount level???[/quote]

I'm serious. In that case, you probably got many problems incorrect and the system feels you need a review.

My problems are way to hard for me. I give up on all of them.  :(

[quote="r15s11z55y89w21"]

You are assigned problems based on how you did on previous problems; it gives you problems that are suitable for your levels.

[/quote]

So if you do really bad, do you keep getting one point questions? In another thread, it says the better you do, the more one point questions you get.

If you are good, and have thus done well on problems, you will get one point problems. If the problems are deemed hard for you, they are worth 4 points. Other problems are 2 points.

Today I logged into Alcumus and did two, but again both of them are 1 pointers (and mathcounts level)... so if 1 pointers mean I have been doing well, then why is it still mathcounts level (on the other hand if it's mathcounts level it implies I've been doing poorly, then why are they still 1 pointers)?

By the way what is the minimum percentage/alcumus score to get aime level problems (if there are) instead of the boring mathcounts level?

Alcumus complements the Introduction to Counting and Probability book/class. All the problems are at that level. If you are competent at that level of math, you will not receive harder problems as they do not exist.

For more information, see the [url=http://www.artofproblemsolving.com/Alcumus/FAQ.php]Alcumus FAQ[/url].


The electrolytic reduction of an organic nitro compound was carried out in a solution buffered by acetic acid and sodium acetate. The reaction was

$ RNO_2 \plus{} 4H_3O^ \plus{} \plus{} 4e \minus{} \minus{} \minus{} > RNHOH \plus{} 5H_2O$

$ 300ml$ of a $ 0.01 M$ solution of $ RNO_2$ buffered initially at $ pH \equal{} 5$ was reduced, with the above reaction going to completion. The total acetate concentration, $ [HOAc] \plus{} [OAc^ \minus{} ]$ was $ 0.50 M$. Calculate the pH of the solution after the reduction is complete.

[hide="IS this correct?"]
$ pH \equal{} pKa \plus{} log(\frac {[OAc^ \minus{} ] \plus{} 4(0.01)}{[HOAc] \minus{} 4(0.01)})$
BUt here I have assumed all $ H^ \plus{}$ comes from $ HOAc$ but it can also come from $ H_2O$ , right?
[/hide]

There was no need to start a new thread when i had already started one.  :mad:
And please don't post short solutions which nobody can understand. Atleast explain your solution... :mad:

What thread? Aren't separate questions need to be posted in separate threads?

Do you know the buffer formula? I have used buffer formula since that is the case. Also the conc. of the salt and acid change because of consumptions of protons. I found the conc. of proton consumed during the reaction and subtracted it from HOAc and added to OAc^- , which gives the correct answer but I am not sure because of the reason I have mentioned in the previous post.Good enough?

I think hell_ever means this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246690


$ f: ( \minus{} 1,1) \rightarrow \mathbb{R}$ and $ f'(0)$ exists. $ x_{n}, y_{n} \rightarrow 0$ as $ n \rightarrow \infty$

Prove that $ \lim_{n \rightarrow \infty} \frac {f(y_{n}) \minus{} f(x_{n})}{y_{n} \minus{} x_{n}} \equal{} f'(0)$ if:

1) $ x_{n} < 0 < y_{n}$

2) $ 0 < x_{n} < y_{n}$ and $ \frac {y_{n}}{y_{n} \minus{} x_{n}} \le M$

For i)

I tried  using what I knew:

$ \lim_{n\ \rightarrow \infty} \frac{f(x_{n})\minus{}f(0)}{x_{n}} \equal{} f'(0)$

and

$ \lim{n \rightarrow \infty} \frac{f(y_{n}) \minus{}f(0)}{y_{n}} \equal{} f'(0)$

then:

$ \lim_{n \rightarrow \infty}\frac{f(y_{n}) \minus{}f(x_{n})}{y_{n}\minus{}x_{n}} \equal{} \lim_{n \rightarrow \infty} \frac{ f(y_{n})\minus{}f(0) \plus{}f(0) \minus{}f(x_{n})}{y_{n} \minus{} x_{n}}$...and then at this point I was just stuck wrangling with the denominator and numerator to no avail....

$ f(t)\equal{}f(0)\plus{}tf'(0)\plus{}tg(t)$, with $ \lim _{t\rightarrow 0}g(t)\equal{} 0$.  We can write $ \frac {f(y)\minus{}f(x)}{y\minus{}x}\minus{}f'(0)\equal{}g(y)\plus{}\frac x{y\minus{}x}(g(y)\minus{}g(x))$
It's easy to see that   [b]"if $ x,y\rightarrow 0 , \ x\neq y$ and $ \frac x{y\minus{}x}$ is bounded"[/b]  then $ \lim _{x,y\rightarrow 0}\frac {f(y)\minus{}f(x)}{y\minus{}x}\minus{}f'(0)\equal{}0$
In our case these conditions are satisfied :
1) $ x_n < 0 < y_n \Longrightarrow |\frac {x_n}{y_n\minus{}x_n}| < 1$

2)$ 0 < x_{n} < y_{n})\& (\frac {y_{n}}{y_{n} \minus{} x_{n}} \le M ) \Longrightarrow 0 <\frac {x_{n}}{y_{n} \minus{} x_{n}}<\frac {y_{n}}{y_{n} \minus{} x_{n}} \leq M$
 :cool:

Why did you use x,y instead of $ x_{n}, y_{n}$?


Let $n \geq 3$ and $k \geq 2$ be integers, and form the forward differences of the members of the sequence
$1,n,n^2,...n^{k-1}$

and successive forward differences thereof, as illustrated on the right for case $(n,k) = (3,5)$. Prove that all entries of the resulting triangles of positive integers are distinct from one another.

Diagram:
http://www.cms.math.ca/Competitions/IMTS/imts5.html

It is easy to see that every number in position (i;j) can be written in a form of (n^i)*((n-1)^j). Thereof, every such number is unique.


If $x,y,z$ are positive reals such that $x^{2006}+y^{2006}+z^{2006}\le 1003$, then $xy+yz+zx\le 4$.
There is a chance this question to be stupid.
Any ideas?

From the power mean ineq and the ineq $x^2+y^2+z^2\geq xy+yz+zx$  we have

$\sqrt[2006] {\frac{1003}{3}}\geq\sqrt{\frac{xy+yz+zx}{3}}$.

Yes , i tought that this maybe something very stupid. The idea wasn`t that but..


It is said in AMC10 booklet that the top 1% will be invited to AIME
what are the usual scores for people who get invited to AIME or top 1%

If I am correct, the AMC will invite either anyone who scored a 120 or higher, or 1 percent of all competitors, whichever is the greater number. In addition, the highest scorer from each state will be invited to take the AIME regardless.

Usually, its 120, but 9th grade, or 2004, it was 110, I think? Because I got invited with a 112ish.

YEAH! That test was harddddd!!

Well... hard for everyone except Yannik.... freaking 146.5   :P

!!!

Contestants who take AMC-10 are invited to AIME if they score at least 120 OR are in 1% of the contestants.

This topic is asked *very* many times and you can also find the information about AMC-12 in the AoPS.


function
calculus
derivative
ratio
calculus computations

This question deals with linear approximations of function values. In particular,

$ f(x) \approx f'(a)\triangle x \plus{} f(a)$ where
$ \triangle x \equal{} dx \equal{} (x\minus{}a)$,
$ \triangle y \equal{} f(x) \minus{} f(a)$
$ dy \equal{} f'(a)\triangle x$

Anyways, since $ \triangle y \approx dy$ then the following apparently holds true:

Suppose that you measure the radius of a circle to be $ 24 \pm 0.2 cm$. Then we have $ A \equal{} 2\pi r^{2} \Longrightarrow A' \equal{} 4\pi(24)r'$. Somehow, we have $ r' \equal{} \pm 0.2$, which I don't understand. How does the derivative of r imply the differential of r? Can someone explain how differentials are used to approximate the relative error of a calculation?

Also, am I off base in saying that $ \frac{dy}{dx} \equal{} \frac{f'(x)\triangle x}{\triangle x} \equal{} f'(x)$ ?

Bump. From the last post, could you say:

$ \frac{dA}{dt} \equal{} 4\pi(24)\frac{dr}{dt} \Longrightarrow dA \equal{} 4\pi(24)dr$ which is approximately $ \triangle A \equal{} 4\pi(24)\triangle r$ ? In other words, you simply cancel the differential of t by multiplying both sides by it and from there the appropriate result follows...?

If we express $ A$ as a function of $ r$, then we have
\[ \frac{\Delta A}{\Delta r}\equal{}A^\prime (s)\]
for some $ s$ between the lowest and highest possible value of $ r$.

When $ \Delta r$ is small, $ A^\prime (s)$ is nearly equal to $ \frac{dA}{dr}$ evaluated at an approximate value of $ r$.

Now the errors in $ A$ and $ r$ are given by $ \frac{\Delta A}{2}$ and $ \frac{\Delta r}{2}$, so the ratio of the errors is also $ A^\prime (s)$, which is nearly equal to $ A^\prime (r)$.


hello,

Find all positive prime $ p$ such that : $ 3^p \minus{} (p\plus{}2)^2\;$ is prime .

That's true, because terms are odd when p is odd and even when p is even. Then the expression is always even, so I guess the answer is $ 2$

[hide="Solution"]$ 3^p\minus{}(p\plus{}2)^2\equiv 1\minus{}p^2\bmod 2$

If $ p\equiv 0\bmod 2$ it's odd, otherwise it's even (and therefore divisible by 2).

Checking, $ p\equal{}2$ gives $ 9\minus{}16\equal{}\minus{}7$. If we're allowing "negative" primes then this works.

Also notice that an even number can be prime (if it's 2). Checking, $ p\equal{}3$ gives $ 2$.

So the only answers are $ 3$ and (possibly) $ 2$.[/hide]


What is the volume of a cube whose surface area is twice that of a cube with volume $ 1$?

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

[hide]
Surface area scales quadratically, so the edge length of the cube is $ \sqrt{2}$.  Thus its volume is $ 2\sqrt{2}$.  Answer is $ C$.
[/hide]

[hide]$ 1^{(1/3)}\equal{}1 \implies 1\cdot6\cdot2\equal{}12 \implies 12/6\equal{}2$ Therefore each side's length is $ \sqrt{2}$. Raising this to the third power we get $ 2\sqrt{2}$. [/hide]

Let $ s$ be a side length of the cube. The surface area of a cube is $ 6s^2$. Then $ 6s^2\equal{}2\cdot 6$, so $ s\equal{}\sqrt{2}$. So the volume is $ s^3\equal{}2\sqrt{2}$, so the answer is $ \boxed{C}$.

If the volume is one, then it would be 6s^2=2*1. This is confusing me.

You're misreading it; the surface area of the first cube is equal to twice the [u]surface area[/u] of the second cube, not twice the volume.

[hide=solution]The surface area of a cube with side length $1$ is $6\cdot 1^{2}=6$. Let $s$ be the side length of a cube with double the surface area. Then $6s^{2}=12\rightarrow s^{2}=2\rightarrow s=\sqrt{2}$. The volume is $\left(\sqrt{2}\right)^{3}=\boxed{\textbf{(C)} ~2\sqrt{2}}$[/hide]


[quote="tranthanhnam"]find all integer positive root of system equation :
$xy+y-x!=1$and $yz+z-y!=1$ and$x^2-2y^2+2x-4y=2$[/quote]

$x^2-2y^2+2x-4y=2$ implies that $y<x$ and then looking at $xy+y-x!=1$ we see that y divides left side so it's also divide 1 hence y=1. This leads to $x=x!$ so $x=2$ (x>y) and $z=1$


[quote="Pain rinnegan"]Let $ a_1$, $ a_2$, ..., $ a_n$ be real nembers st
a) $ a_1\leq a_2\leq ...\leq a_n$
b) $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 0$
c) $ \left|a_1\right| \plus{} \left|a_2\right| \plus{} ... \plus{} \left|a_n\right| \equal{} S$.
Prove that $ a_n \minus{} a_1 \geq \frac {2S}{n}$.[/quote]
If $ S\equal{}0$ it's obvious.
Let $ S>0$ and $ a_1\leq...\leq a_k\leq0\leq a_{k \plus{} 1}\leq...\leq a_n.$
Hence, $ \frac {S}{2} \equal{} \minus{} \sum_{i \equal{} 1}^ka_i\leq \minus{} ka_1$ and $ \frac {S}{2} \equal{} \sum_{i \equal{} k \plus{} 1}^na_i\leq(n \minus{} k)a_n.$
Thus, $ a_n \minus{} a_1\geq\frac {S}{2}\left(\frac {1}{k} \plus{} \frac {1}{n \minus{} k}\right)\geq\frac {2S}{n}.$


When is the textbook going to come out?

In WOOT, I saw that two of the topics are Number Theory.

In Woot, does it go into that much detail?

THanks!

The book is on the "do eventually" list.  We don't have a release date for it.

I'm not sure how to describe how "in detail" the WOOT classes are -- they introduce concepts in number theory and give you some practice problems.  They are not a full course in number theory.


A simplified model of a bicycle of mass $ M$ has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is $ w$, and the center of mass of the bicycle is located midway between the tires and a height $ h$ above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration has a magnitude $ a$. Air resistance may be ignored. Assume that the coefficient of sliding friction between each tire and the ground is different: $ \mu_1$ for the front tire and $ \mu_2$ for the rear tire. Let $ \mu_1\equal{}2\mu_2$. Assume that both tires are skidding: sliding without rotating. What is the maximum value of $ a$ so that both tires remain in contact with the ground?

Additionally, I'm confused how this case is different from the case that $ \mu_1\equal{}\mu_2$.

I know it now. :D  :lol:
One thing I forgot to say in the problem,the answer should be expressed in terms of $ w$, $ h$, $ M$, and $ g$.Sorry.

[hide="Now you can explain why, simon_math"]$ \mu_2g<a<\mu_1g$[/hide]

Well,I'm confused once more. I just thought the two cases($ \mu_1 \equal{} \mu_2$ and $ \mu_1 \equal{} 2\mu_2$) are equal, and I know the answer when $ \mu_1 \equal{} \mu_2$, so I thought I could solve this case as well. But I think you mean they are different.Why do you think so? My reason is that when the rear wheel is leaving the ground, there's no frction acting on that wheel, so we don't have to care about the difference in $ \mu_1$ and $ \mu_2$ any more.
With the balance of torques, I got $ a \equal{} \frac {wg}{2h}$

Actually, $ a < {{wg}\over{2h}}$, otherwise the rear wheel (or tire), will loose contact with the ground.
But, if  $ \mu_1 > \mu_2$  ,as in the first post version, and $ \mu_1 < {w\over{2g}}  \Rightarrow \mu_2 < {a\over g} < \mu_1$, (the left inequality in order to avoid the hop of the front wheel).

If  $ \; \mu_1 \equal{} \mu_2 \equal{} \mu,\;\; a \equal{} \mu g \Rightarrow \mu < {w \over {2h}}$.


In the plane $Oxy$, two points $A \in xOx'$ and $b \in yOy'$ move and always satisfy the condition $AB=3$. Given $J \in AB$ with $\vec{AJ}=2\vec{JB}$.
Find locus of point $J$.

Let's consider this locus only for the upperhalf plane.Then if $A = (t,0)$, we have \[B = (0, \sqrt{9-t^{2}}).\] It means that \[\vec{OA}= \left[t, 0 \right]\] and \[\vec{OB}= \left[0, \sqrt{9-t^{2}}\right],\] so \[\vec{OJ}= \frac{1}{3}\vec{OA}+\frac{2}{3}\vec{OB}= \left[ \frac{1}{3}t,\frac{2}{3}\sqrt{9-t^{2}}\right].\] Substituting $x=\frac{1}{3}t$ we get \[\vec{OJ}= \left[ x,\frac{2}{3}\sqrt{9-9x^{2}}\right]\iff J=\left( x,\frac{2}{3}\sqrt{9-9x^{2}}\right).\] It means that this locus in the upperhalf plane is graph of function \[f(x)=2\sqrt{1-x^{2}}\] (for lowerhalf it's symmetric), so in fact this is an ellipse.


inequalities
Ramsey Theory
combinatorics unsolved
combinatorics

does anybody know a simple, i mean a non-computer calculations invlolved proof that the Ramsey number R(3,7) <= 23 ? thanks in advance to everyone who replies.
I've tried some googlization for it but unsuccessfully so far. I was able to find precise uper bounds for R(3,8) for example, but much computations were used to achieve the result.

It is almost certain that any proof of this fact is ugly. The standard proof gets you $R(3,7)\le R(2,7)+R(3,6)=7+R(3,6)$, and continuing down the chain $R(3,7)\le 28$.

The problem is that in general, exact calculations of Ramsey numbers are very ugly, unless there is some amazing and highly symmetric graph which makes the standard bound exact.

yes, i agree with you. Considering that R(3,6) = 18 (and one can show that R(3,6) <= 18 without any help of computers !) you get that R(3,7) <= 25. However, there can hardly exist any approach to proving that the exact value is 23  without massive computations :(. I asked because i wondered whether anyone in the forum knew about a non-computer aided proof of R(3,7) <= 23 since such exists for the inequality R(3,6) <= 18. I still hope that this could be not impossible.


find all  x,y,z,
such:
xy+1&xz+1&yz+1 are perfect squares.

Perhaps he want to say :
[quote="dashmiz"](...) such that xy+1, xz+1 and yz+1 are perfect squares.[/quote]

I don't think it's possible to find [b]all[/b] $(x,y,z)\in\mathbb{N}^3$ such that $xy+1$, $xz+1$ and $yz+1$ are perfect squares. But we can find infinitely many solutions with: $(x,y,z)=(k-1,k+1,4k)$, $k\in\mathbb{N}^*$.
Have a look at: [url=http://www.math.hr/~duje/dtuples.html]diophantine d-tuples[/url]

yes xy+1

and   xz+1

and   yz+1

are perfect squares.


LaTeX
induction
trigonometry
limit
algebra unsolved
algebra

Given {b_n} is sequence satisfy

b_1 = 1/2

b_{n+1} = 1/2 ( b_ n + sqrt{(b_n)^2 + 1/(4^n) } )

find lim b_n

P/S : anyone please edit it for me . I am not be able to type latex right now .

[quote="HTA"]Given $ \{b_n\}$ is sequence satisfy

$ b_1 \equal{} \frac 12$ and $ b_{n\plus{}1} \equal{} \frac{ b_ n \plus{} \sqrt{(b_n)^2 \plus{} \frac 1{4^n} } }2$

find $ \lim_{n\to \plus{}\infty} b_n$.[/quote]

[hide="Solution"]
It's easy to show with induction than $ b_n\equal{}\frac{1}{2^n\tan(\frac{\pi}{2^{n\plus{}1}})}$

And so $ \lim_{n\to\plus{}\infty}b_n\equal{}\frac 2{\pi}$
[/hide]


geometry
probability
rotation
analytic geometry

Mandelbrot 2005 October Round 1

Choose a point $O$ at random inside a given circle with radius $3$, and also select a real number $r$ at random between $0$ and $2$. What is the probability that the circle with center $O$ and radius $r$ intersects the original circle?

[hide]
WLOG rotate $O$ around the origin so it lies on the x-axis.  Let the x-coordinate of $O$ be $x$, and $r = y$.  Then we can graph $0 < x < 3$, $0 < y < 2$, $x + y > 3$.  Graphing this, we find the satisfied area has area $2$, and the total area is $6$, giving the probability $\frac 1 3$
[/hide]

the lowest value of r that intersects the big circle is the one that makes it tanget.

let q be the distance of point o from the center of the big circle
then 2>r>3-q
if you plot r and q you get a triangle with legs of 2 and a rectanlge 2 by 3
the area of the triangle is 2*2*.5 so the probabilty is 2/6 or 1 third.


The solutions to $8t^{3}+4t^{2}-4t-1=0$ are $\cos{2\pi\over 7},\cos{4\pi\over 7},\cos{6\pi\over 7}$ (known result), hence

$a_{n}=A\cos^{n}{2\pi\over 7}+B\cos^{n}{4\pi\over 7}+C\cos^{n}{6\pi\over 7}$

From the initial conditions we get $A=B=C=1$ (I will leave the proof to someone else), therefore

$a_{n}=\cos^{n}{2\pi\over 7}+\cos^{n}{4\pi\over 7}+\cos^{n}{6\pi\over 7}$

[quote="SplashD"][quote="Farenhajt"]The solutions to $8t^{3}+4t^{2}-4t-1=0$ are $\cos{2\pi\over 7},\cos{4\pi\over 7},\cos{6\pi\over 7}$ (known result)[/quote]

i was just wondering,

what obscure things do people memorize/know well for contests, like the solutions of this polynomial?[/quote]

you don't usually memorize these things, but if you do a problem concerning that idea, you should be able to recognize that situation in another problem


[url]http://www.cafepress.com/fractalfort[/url]

Tell me what you think!

It would be nice if you had some t-shirts in other colors than white. Also, you may want to gather some more jokes.


Hola. Se que no es muy matematico el post pero puede ser de interes para algunos latinoamericanos. Resulta que mañana los colegios alemanes de latinoamerica daran esta prueba de idioma y me gustaria contactarme con alguna otra victima. Mi msn es pasten[arroba]matematicas[punto]net (lo escribi asi para evitar mas spam) y mi e-mail es hpasten[arroba]gmail[punto]com
Saludos y gracias.

Hola... a mi me toco presentar el Sprachdiplom 1 hace un a~no y medio. En el colegio nos decian que era un examen muy dificil y le hablan a uno una cantidad de carreta que lo unico que hace es asustarlo a uno. El examen es facil, yo que soy muy malo en aleman lo pase bien. Entonces no es para preocuparse. Si es el 2 no se, a mi por el calendario del colegio me toca en noviembre o algo asi, no se bien.

[color=blue]Yo también estudio alemán en mi colegio. El año pasado, en el mes de Septiembre, me correspondió hacer a prueba del DSD I. El próximo año, en Agosto, creo, tendré que hacer el DSD II.

El DSD I no es muy difícil, salvo, tal vez, por el Lückentext. El DSD II parece ser más fuerte, pero nos estuvieron preparando todo este año para esta prueba, y ya domino algunas de las técnicas que suelen pedir para dichos exámenes.

Lo que no da gusto del alemán de mi colegio es que sólo tuve una clase de vocabulario matemático en dicho idioma. No es en el año, sino en toda la vida. Fue entre dediados y finales… ¡del año 2000! ¿No es cierto que así no da gusto?[/color]


inequalities
integration
number theory
least common multiple
inequalities unsolved

I think that problem is very nice, in my opinion quite non-standard:
Próve that:
$ \forall_{n\in \mathbb{N}} SCM(1,2,...,n)\geq 2^{n\minus{}1}$
:wink:


geometry
circumcircle
Euler
trigonometry
geometry unsolved

Let $ K$ be a point which is symmetric to $ O$-the circumcenter of triangle $ ABC$,due to side $ BC$.Prove that the Euler line
of triangle $ ABC$ intersects the side $ AK$ at its midpoint.

Call H the orthocenter, M the midpoint of BC and N of AH. Call F the center of the Feuerbach circle of ABC, then MF=FN and $ AN\equal{}b \frac{\cos A}{\sin B} \equal{} R \cos A \equal{} MK$ so $ \triangle MKF$ and $ \triangle AFN$ are equal and AK pass tought F and $ AF\equal{}FK$.


ratio
trigonometry
inequalities proposed
inequalities

Find the smallest value of $ \alpha$ such that:For all triangles $ ABC$ we can choose two sides $ a,b$ satisfying
\[ 1\le \frac {a}{b} < \alpha
\]
@My English is very bad,so if you don't understand my problem ,I will try to explain it  :)

Assume without loss of generality the largest side is $ 1$, another side is $ a$, and express the third side by the law of cosines.  The smallest possible ratio of two sides is,
\[ r(a,\theta) \equal{} \min \left \{\frac {1}{a},\frac {1}{\sqrt {1 \plus{} a^2 \minus{} 2a\cos \theta}},\max \left \{\frac {a}{\sqrt {1 \plus{} a^2 \minus{} 2a\cos \theta}},\frac {\sqrt {1 \plus{} a^2 \minus{} 2a\cos \theta}}{a}\right\}\right \}
\]
then argue that the maximum must be achieved at $ \theta \equal{} 0$, so that the problem is reduced to finding the maximum of
\[ r(a) \equal{} \min \left \{\frac {1}{a},\frac {1}{1 \minus{} a},\max \left \{\frac {a}{1 \minus{} a},\frac {1 \minus{} a}{a}\right\}\right \}
\]
for $ a\in (0,1)$, which is fairly straightforward.
  [hide="the answer comes out to"]$ \alpha \equal{} \max _{a\in (0,1)}r(a) \equal{}\frac{1\plus{}\sqrt{5}}{2}$.[/hide]

Since $ \frac a{b} \ge 1$, there are 2 cases to consider:
(1) $ a\ge b\ge c$ and $ \frac a{b} \le \frac b{c}$;
(2) $ c\ge a\ge b$ and $ \frac a{b} \le \frac c{a}$.
In (1), from $ a<b\plus{}c$ we get $ \frac a{b} < \frac b{a\minus{}b}$, therefore $ \frac a{b}< \frac {1\plus{}\sqrt 5}{2}$.
In (2), from $ c<a\plus{}b$ we get $ \frac a{b}<\frac {a\plus{}b}{a}$, therefore $ \frac a{b}< \frac {1\plus{}\sqrt 5}{2}$.

The following is also interesting.

Find the smallest value of $ \alpha$ such that:For all quadrilaterals  ABCD we can choose two sides  $ a$ and $ b$ satisfying

$ 1\le \frac a{b} <\alpha$.


vector
function
Functional Analysis
advanced fields
advanced fields unsolved

Let $M$ be a normed (Bahach) vector space. And let $f\colon M\to M$ be an isometry, that is $\|f(x)\|=\|x\|$ for all $x\in M$. Does this imply that $f$ is linear?

Note that it is obviously true when $M$ is a eucledean space with the norm induced by the inner product.

No, it does not.

Here is a hint how to construct an example.
1) Let $f:{\mathbb R}\to L_{1}({\mathbb R})$ be $x\mapsto I([0,x])$ be the indicator of the interval $[0, x]$. Prove that $f$ has the properties you want (except that it maps one normed space to a different normed space).
2) Extend $f$ to a mapping from ${\mathbb R}\oplus L_{1}({\mathbb R})$ to $L_{1}({\mathbb R})\oplus L_{1}({\mathbb R})$.
3) Prove that both ${\mathbb R}\oplus L_{1}({\mathbb R})$ and $L_{1}({\mathbb R})\oplus L_{1}({\mathbb R})$ are isomorphic to $L_{1}({\mathbb R})$ (we define the norm in the direct sum of spaces as follows $\|(a,b)\| =\|a\|+\|b\|$ ).

Conclude that there exists the desired mapping from $L_{1}({\mathbb R})$ to itself.

BTW, this mapping is an isometry of metric spaces ($L_{1}({\mathbb R})$ and its subset): \[\|f(x)-f(y)\| = \|x-y\|.\] This property is stronger than that you wanted.

Thanks, Yury. Your example is very nice. Sorry, i have realized that i had posted something different from that i realy had wanted. I know it is a bad behaviour to correct problems on the way, so please forgive me :)

[i]The new version of the problem.[/i]
Let $M$ be a Banach space and $f\colon M\to M$ be a function such that
(a) $\|f(x)-f(y)\|=\|x-y\|$ for all $x,y\in M$,
(b) $f(0)=0$,
(c) $f$ is a bijection.

Prove or disprove that $f$ is linear.

It suffices for $M$ to be a normed real space (completeness plays no role here).

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=isometry&t=71982]this[/url]. I know it's not exactly a complete proof, but it might shed some light on what's going on, and I think it can be made into a proper solution.


geometry
rotation
geometric transformation
calculus
calculus computations

Consider the curve x=-t^2, y=[(t^3)/3]-t+1 , 0<t<1
Find the area of the surface obtained by revolving the curve about the line y=x.

How to do revolving a curve around non-axis line???

OR should I tranform the whole thing such that it'll become revolving about x-axis??
ie) Let x=-t^2 and y'= y-x = {[(t^3)/3]-t+1}- {(-t^2)} , revolving about x-axis?

You're trying to skew, but skewing changes the surface area.  Instead, try to rotate your system to get an axis you like.  (Actually, if you think about how to prove the formulas for surface areas of solids of rotation are, you should be able to figure out what the correct version is for rotating around lines that aren't parallel to one of the axes without explicitly rotating.)


conics
ellipse
geometry unsolved
geometry

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14755[/img]

 :(

See [url]http://www.mathlinks.ro/viewtopic.php?t=209729[/url].

Please do not double post.


Αν $ a_{1},a_{2},...,a_{n}$ είναι θετικοί αριθμοί με $ n>\equal{}3$ να αποδειχθεί ότι $ \frac{a_{1}}{a_{2}\plus{}a_{3}\plus{}...\plus{}a_{n}}\plus{}\frac{a_{2}}{a_{1}\plus{}a_{3}\plus{}...\plus{}a_{n}}\plus{}...\plus{}\frac{a_{n}}{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n\minus{}1}}\geq \frac{n}{n\minus{}1}$

[hide]θέτουμε $ S\equal{}\sum^{n}_{i\equal{}1}a_{i}$ για διευκόλυνση[/hide]

[hide]Ανισότητα της αναδιάταξης[/hide]

Hmm, δεν το προσπάθησα έτσι. Here is my solution....

Εαν θέσουμε με $ S\equal{}\sum^{n}_{i\equal{}1}a_{i}$, τότε η ανισότητα παίρνει τη μορφή $ \sum^{n}_{i\equal{}1}\frac{a_{i}}{S\minus{}a_{i}}\geq \frac{n}{n\minus{}1}$. Όμως

$ \sum^{n}_{i\equal{}1}\frac{a_{i}}{S\minus{}a_{i}}\equal{}\sum^{n}_{i\equal{}1}\frac{a_{i}\minus{}S\plus{}S}{S\minus{}a_{i}}\equal{}S(\sum^{n}_{i\equal{}1}\frac{1}{S\minus{}a_{i}})\minus{}n$ (1)

Όμως από την ανισότητα του Cauchy έχουμε ότι $ S\sum^{n}_{i\equal{}1}\frac{1}{S\minus{}a_{i}}\geq \frac{nS}{\sqrt[n]{\prod^{n}_{i\equal{}1}(S\minus{}a_{i}})}$.  (2)

Από την ανισότητα του Cauchy παίρνουμε επίσης ότι $ \frac{n}{\sqrt[n]{\prod^{n}_{i\equal{}1}(S\minus{}a_{i})}}\geq \frac{n^{2}}{(n\minus{}1)S}$   (3)

Απο τις σχέσεις (2) και (3) παίρνουμε ότι $ S\sum^{n}_{i\equal{}1}\frac{1}{S\minus{}a_{i}}\geq \frac{n^{2}}{n\minus{}1}$

Τελικά, από τη σχέση (1) προκύπτει ότι $ \sum^{n}_{i\equal{}1}\frac{a_{i}}{S\minus{}a_{i}}\geq \frac{n^{2}}{n\minus{}1}\minus{}n\Longleftrightarrow \sum^{n}_{i\equal{}1}\frac{a_{i}}{S\minus{}a_{i}}\geq \frac{n}{n\minus{}1}$ :wink:

Ναι για Θάλη Β' Γυμνασίου είναι :P

Δείτε εδώ:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130052&start=20

Τη 16 :)

[quote="mostel"]Ναι για Θάλη Β' Γυμνασίου είναι :P[/quote]

Ρε mostel όταν λέω junior level εννοώ ότι είναι από βαλκανιάδα, μόνο που εκεί έδιναν κάτι παραπάνω στα παιδιά, ότι δηλαδή $ \sum^{n}_{i \equal{} 1}a_{i} \equal{} 1$ :D

Α! και η λύση του Κώστα είναι πολύ κομψή( Κώστα μη στηρίζεσαι όμως πολύ στον Andreescu!! :!: )

Βγαίνει και με Jensen αλλά όπως λέει και ο κύριος Λουρίδας ασ μην ψάχνουμε για φαντάσματα... :lol:

xwris vlavi tis genikotita ipothetw $ a1 >\equal{} a2 >\equal{} a3 >\equal{}.....>\equal{}an$  . exw omws
$ 1/(a2\plus{} a3 \plus{}.. an) >\equal{} 1/(a1 \plus{} a3 \plus{} ... \plus{} an) >\equal{} ... >\equal{}  1/(a1\plus{} a2 \plus{}...\plus{} an\minus{}1)$
ara afou oi niades exoun omoia diataksi apo thn anisotita tis anadiataksis exw:
$ a1/(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} a2/(a1\plus{}a3\plus{}...an) \plus{} .....\plus{}an\minus{}1/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an/(a1 \plus{} a2 \plus{}...an\minus{}1 ) >\equal{} a2 / ( a1 \plus{} a3 \plus{} ... \plus{} an) \plus{}   a1/(a2 \plus{} a3 \plus{}...an) \plus{}....\plus{} an/(a1 \plus{}...\plus{} an\minus{}2 \plus{} an) \plus{} an\minus{}1/(a1 \plus{}...\plus{}an\minus{}1)$

episis exw
$ a1/(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} a2/(a1\plus{}a3\plus{}...an) \plus{} .....\plus{}an\minus{}1/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an/(a1 \plus{} a2 \plus{}...an\minus{}1 ) >\equal{} a3/(a2 \plus{} a3 \plus{}... \plus{} an) \plus{} a4/(a1 \plus{} a3 \plus{}.... \plus{} an) \plus{}.....\plus{} an\minus{}3/ (a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an\minus{}1/(a1 \plus{} a2 \plus{}...an\minus{}1 )$
omoia prattontas loipon sximatizw $ (n\minus{}1)$ sxeseis simfwna me tn anisotita tis anadiataksis oi opoies me prosthesi kata meli dinoune:
$ (n\minus{}1) * (  a1/(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} a2/(a1\plus{}a3\plus{}...an) \plus{} .....\plus{}an\minus{}1/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an/(a1 \plus{} a2 \plus{}...an\minus{}1 ) ) >\equal{} ( a2 \plus{} a3 \plus{}... an) /(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} (a1\plus{}a3\plus{}...an)/(a1\plus{}a3\plus{}...an) \plus{}....\plus{} (a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an)/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} (a1 \plus{} a2 \plus{}...an\minus{}1 )/(a1 \plus{} a2 \plus{}...an\minus{}1 )$
 <=>
$ (n\minus{}1)*(  a1/(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} a2/(a1\plus{}a3\plus{}...an) \plus{} .....\plus{}an\minus{}1/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an/(a1 \plus{} a2 \plus{}...an\minus{}1 ) ) >\equal{} (1\plus{}....\plus{}1)$

<=>
$ (n\minus{}1)*(  a1/(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} a2/(a1\plus{}a3\plus{}...an) \plus{} .....\plus{}an\minus{}1/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an/(a1 \plus{} a2 \plus{}...an\minus{}1) ) >\equal{} n*1$

<=>
 $ a1/(a2 \plus{} a3 \plus{}.... \plus{}an) \plus{} a2/(a1\plus{}a3\plus{}...an) \plus{} .....\plus{}an\minus{}1/(a1 \plus{} a2 \plus{}..\plus{}an\minus{}2 \plus{} an) \plus{} an/(a1 \plus{} a2 \plus{}...an\minus{}1) >\equal{} n/(n\minus{}1)$,   poy einai k to zitoumeno.

ayti einai i lysi mou. an kapoios paratirisei kati p pisteyei oti ekana lathos na m to pei  :)  opws kai gia kapoia dieykrinisi
P.S. eyxaristw ton protonio gia tn latex  :lol:


function
LaTeX
real analysis
real analysis unsolved

Hi all,

Assume f(x) and g(y) two functions both in L^2(R),
a separable function is defined as (x,y) -> f(x)g(y), which is in L^2(R^2)

I want to know if the set of these separable functions is dense in L^2(R^2).

Intuitively, I think not but I can not give a proof. So can anyone help me ?

Thanks.

[quote="Kent Merryfield"]These separable functions are dense in $L^2(\mathbb{R}^2).$
[/quote]
Just a minor remark: what Kent proved is that the linear span of separable functions is dense (a very useful fact that, most likely, was meant), not that the set of separable functions itself is dense (which is what the problem literally asked about and which is not true). I'm writing this just to prevent any possible confusion  :) .

fedja - that's exactly why I deleted my post. I realized that that's all I had proved.

Here's an example which cannot be closely approximated by separable functions- you should be able to build a proof around it.

$f(x,y)=1$ if $-1<x<0$ and $0<y<1$
$f(x,y)=1$ if $0<x<1$ and $-1<y<0$
$f(x,y)=0$ otherwise.

Nice example.

Thanks all for the kind responses. :)

btw: I just became a new member here. So how can u enter
        those Latex-style equations in the posts ? Thanks.

$\LaTeX$ math mode is enabled on this forum. Just put your code inside dollar signs.


\[ \dfrac{x}{\sqrt{y^2+z^2}} \ge \dfrac{2x^2}{x^2+y^2+z^2} \Leftrightarrow \]
\[ x^2+y^2+z^2 \ge 2x \sqrt{y^2+z^2} \]
which is true by AM-GM.
Therefore
\[ \dfrac{x}{\sqrt{y^2+z^2}}+\dfrac{y}{\sqrt{z^2+x^2}}+\dfrac{z}{\sqrt{x^2+y^2}} \ge \dfrac{2x^2+2y^2+2z^2}{x^2+y^2+z^2}=2. \]
As the equality holds iff $x^2=y^2+z^2,y^2=z^2+x^2,z^2=x^2+y^2$, which is impossible, so the inequality is strict.

this inequality was the subject of a test in 1994, if i'm not mistaken. :?

[quote="Detective"]If
$x=1, y=1, z=0$ or its permutation,

we get
$2\geq2$

so equality can occur.

thus the inequality is not strict.

am i right? or did i miss something?[/quote]

x,y,z can't be 0. You can only take them to be tending to 0. So 2 is an infimum not minimum.


My school is registrating for next year's classes tomorrow.  I will take ap calc, ap us history,ap lit, ap chem.  I want to take AP euro history, but i don't know if i can handle all that ap classes.  How hard is the euro history test? should i take it?
Thanks

i took AP global history, which is sortof similar but also different from the AP euro class, which my brother is currently enrolled in. i feel euro places alot of emphasis on specifics of european monarchs- which may or may not be a good thing depending on your style of learning.
you have to keep in mind though, with ap lit, american and euro you may get alot of reading or essays that have to be done all at once- which may not be a most pleasant thing to go through. if you're good with time management, i'd say go for it... you may want to take a look at the syllabus on collegeboard.com

wow, sounds like you might start looking like your avatar, certainly i would with that many AP classes, heh

hmmm I'm taking APUSH, AP Calc AB, AP Chem, and AP English III.  I do not like the history AP's, too much reading and such.  They take more time than all the others.

[quote="runningking7"]I'm taking AP US, AP Euro, and AP Calc now, and let me tell you, do not ever take two history classes simultaneously unless you're a fast reader. Unfortunately in my school, we are required to take TONS of notes everyday, which add up to four hours per day, actually. Find out your teacher's policies first and his/her grading scale.[/quote]

really? our school would never let us take 2  history classes in one year, its just too wierd...unless u have ap world and ap euro cause they kinda overlap. what grade are u in?

i will be a junior next year.
i have decided not to take ap euro, way too much work, instead i signed up for psychology/socialogy (can't spell that word ;) )


[quote="mecrazywong"]Why I cannot attach files in some forums?[/quote]

I dunno...

I thought you could add them everywhere here, but you can't add any type of file. Only files that we think can be useful. (like you can't store video files etc.)

Here is some examples:
AoPS - MathLinks Forum Index -> International Contests & Programs -> Asia
AoPS - MathLinks Forum Index -> International Contests & Programs -> Europe
AoPS - MathLinks Forum Index -> International Contests & Programs -> Africa & Australia

I think normal users cannot attach files in these forums.

Should be fixed for those forums now (i.e. you should be able to attach files now).

if ever necessary, you can also attach them elsewhere and link to the file into another forum...

Cool idea about videofilms, Peter! I'm already trying to imagine people posting LOtR DVDs on MathLinks...

  darij

Hmm... yeah would be cool... 10 gigs... 2 MB per person... you'd need to make 5000 accounts though :D


videos
number theory
relatively prime
SFFT
special factorizations

[hide]if $\frac{1}{m}+\frac{1}{n}=\frac{2}{15}$, then $\frac{m+n}{mn}=\frac{2}{15}$, or $15m+15n=2mn$. So, $2mn-15m-15n=0$. Or, $4mn-30m-30n=0$. Then, by Simon's favorite factoring trick $(2m-15)(2n-15)=225$.

So the factors of of 225 are either 15, 1 and 225, 5 and 45, 3 or 75, or 9 and 25

We want the factors to be relatively prime so our choices are 1 and 225, or 9 and 25

Let's start with the factors being 1 and 225. So $2m-15=225$ and $2n-15=1$, so $m=120$ and $n=8$, also $n$ does divide $m$,

Quickly checking our other case produces a $n$ that does not divide $m$, $2m-15=25$ and $2n-15=9$, so $m=20$ and $n=12$. So the sum is 32

So the sum is $32$

[/hide]

edit: I messed up...my bad

[quote="slamchess"][quote="stupidityismygam"][hide]if $\frac{1}{m}+\frac{1}{n}=\frac{2}{15}$, then $\frac{m+n}{mn}=\frac{2}{15}$, or $15m+15n=2mn$. So, $2mn-15m-15n=0$. Or, $4mn-30m-30n=0$. Then, by Simon's favorite factoring trick $(2m-15)(2n-15)=225$.

So the factors of of 225 are either 15, 1 and 225, 5 and 45, 3 or 75, or 9 and 25

We want the factors to be relatively prime so our choices are 1 and 225, or 9 and 25

Let's start with the factors being 1 and 225. So $2m-15=225$ and $2n-15=1$, so $m=125$ and $n=8$, also $n$ does not divide $m$, so this is a solution. So the sum is 133.

Quickly checking our other case produces a $n$ that divides $m$

So the sum is $133$

[/hide][/quote]

wat is simons favorite factoring trick???[/quote]

there is a video on aops wiki, but the basic use for it is

if you have like $xy+x+y=0$, then in order to get this into factorable form, you add $1$ to both sides, so that you get $xy+x+y+1=1$ or $(x+1)(y+1)=1$. This makes it easier to find solutions for $x,y$


Consider the following sequence:
$x_1=1$ and $x_{n+1}=x_n + \frac{1}{x_n}$.

Prove that $x_{100} > 14$


The perimeter of a certain right triangle is an integer. If the difference between the lengths of the hypotenuse and the longer leg is the same as the difference between the lengths of the two legs, what is the least possible perimeter of this triangle?

Source: another MD Math Team team round question

Is something wrong with my reasoning? I got that therefore, the short leg and the hypotenuse are the same length. However, the hypotenuse is always the longest, but in this case one of the legs is. So therefore, no such triangle. Am I doing something wrong?

how did you get that the shorter leg and the hypotenuse are the same lengths?

Perhaps you made a mistake somewhere?

[hide="HINT"]use pythagorean (sp?) triples[/hide]

Whoops! Found my mistake. Let's call the hypotenuse h, the shorter leg x, and the longer one y.
h - y = y - x
h = 2y - x

My bad  :)

I accidentally set up the equation as h - y = x - y.