inequalities
limit
inequalities proposed

I have a math problem. Can you help me. Prove $ 1\plus{}1\equal{}2$ :D.
P/s:I don`t mad

by $ AM\minus{}GM$, we have
$ 1\plus{}1\ge 2$
$ 1\equal{}1\equal{}>.....$
 :rotfl:

[quote="Nguyen Hoang Nam"]I have a math problem. Can you help me. Prove $ 1 \plus{} 1 \equal{} 2$ :D.
P/s:I don`t mad[/quote]
we have:
$ 1\equal{}1 <\equal{}> (1\minus{}1)^2\equal{}0 <\equal{}> 1\plus{}1\minus{}2\equal{}0 <\equal{}> 1\plus{}1\equal{}2$  :rotfl:  :rotfl:  :rotfl:
[hide="To VietNamese members "]híc,mấy thằng ku này định phá diễn đàn đấy hả? :rotfl:  :rotfl: =>coi chừng bị ban nick đấy  :lol: [/hide]

[quote="hophinhan"]2 cậu làm ơn, bỏ cờ Việt Nam xuống dùm tui cái...chắc lại là mấy đứa nhock từ diendan3t ra  :mad:

Mod plz del this topic  :blush:[/quote]

sr Nhạn, 2 people use
let $ a,b,c\ge 0$
CMR ([b]prove that[/b]) thì phải
$ \sum (\frac{a}{b\plus{}c})^3 \plus{}\frac{9abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq \sum \frac{a}{b\plus{}c}$

$ x\equal{}\frac{2a}{b\plus{}c}$
$ y\equal{}\frac{2b}{c\plus{}a}$
$ z\equal{}\frac{2c}{a\plus{}b}$

$ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$
..............................
please help me by pqr

[quote="Nguyen Hoang Nam"]I have a math problem. Can you help me. Prove $ 1 \plus{} 1 \equal{} 2$ :D.
P/s:I don`t mad[/quote]
 If you have $ 1 \plus{} 1 \equal{} 2$ it is very interesting
Example $ 2^n \equal{} (1 \plus{} 1)^n \equal{} \sum_{k \equal{} 0}^{n}{C_{n}^{k}}$
then Bunhiacopxki inequality we have $ 4^n \equal{} (\sum_{k \equal{} 0}^{n}{C_{n}^{k}})^2 \leq (n \plus{} 1) \sum_{k \equal{} 0}^{n}{(C_{n}^{k})^2}$
then $ \sum_{k \equal{} 0}^{n}{(C_{n}^{k})^2} \geq \frac {4^n}{n \plus{} 1}$
 Or $ ln2 \equal{} ln(1 \plus{} 1) \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}{\frac {( \minus{} 1)^{k \minus{} 1}}{k}}$ If you want find $ \lim_{n \to \plus{} \infty} {\sum_{k \equal{} 1}^{n}{\frac {( \minus{} 1)^{k \minus{} 1}}{k}}}$ ...................  :play_ball:

[quote="fesha3t"][quote="hophinhan"]2 cậu làm ơn, bỏ cờ Việt Nam xuống dùm tui cái...chắc lại là mấy đứa nhock từ diendan3t ra  :mad:

Mod plz del this topic  :blush:[/quote]

sr Nhạn, 2 people use
let $ a,b,c\ge 0$
CMR ([b]prove that[/b]) thì phải
$ \sum (\frac {a}{b \plus{} c})^3 \plus{} \frac {9abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq \sum \frac {a}{b \plus{} c}$

$ x \equal{} \frac {2a}{b \plus{} c}$
$ y \equal{} \frac {2b}{c \plus{} a}$
$ z \equal{} \frac {2c}{a \plus{} b}$

$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4$
..............................
please help me by pqr[/quote]

q=xy+yz+zx
p=x+y+z
r=xyz
my solution
$ \sum (\frac {a}{b \plus{} c})^3 \plus{} \frac {9abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq \sum \frac {a}{b \plus{} c}$
$ <\equal{}> p^3\plus{}12r\plus{}3pr\geq 16p$
$ <\equal{}> p^3\plus{}(3r\minus{}16)p\plus{}12r\geq 0$

$ f(p)\equal{}p^3\plus{}(3r\minus{}16)p\plus{}12r$
$ f'(p)\equal{}3p^2\plus{}3r\minus{}16$
$ f'(p)\equal{}0<\equal{}>p\equal{}\sqrt{(16\minus{}3r)/3}\ge \sqrt{13/3}$
$ \equal{}> f'(p)>0 with p\in [3;4]$
$ \equal{}> Min f(p)\equal{}min f(3)$

If $ p\geq 4$
we have$ r\geq \frac{16p\minus{}p^3}{4p\plus{}9}$

[quote="Nguyen Hoang Nam"][quote="hophinhan"]2 cậu làm ơn, bỏ cờ Việt Nam xuống dùm tui cái...chắc lại là mấy đứa nhock từ diendan3t ra  :mad:

Mod plz del this topic  :blush:[/quote]

I don`t think so you  :rotfl:  :rotfl:  :rotfl:  :rotfl: . And I am Nam`s brother. If you learn in university, you can learn very many math branch. You can learn so much of this problem. Bye
Nam`s brother[/quote]
Can you please show me how to prove that $ 1 \plus{} 1 \equal{} 2$?  :oops:

I've tried some of the approaches. And $ (1 \minus{} 1)^2 \equal{} (1 \minus{} 1)\cdot(1 \minus{} 1) \equal{} 1 \cdot 1 \plus{} ( \minus{} 1) \cdot 1 \plus{} 1 \cdot ( \minus{} 1) \plus{} ( \minus{} 1) \cdot ( \minus{} 1) \equal{} 1 \plus{} 1 \minus{} 1 \minus{} 1$, but i don't see how that implies $ 1 \plus{} 1 \equal{} 2$ :S

Edit: I looked it up, and to solve this problem we need Peano Postulates plus some definitions:

[b]P1.[/b] $ 1 \in \mathbb{N}$
[b]P2.[/b] If $ x \in \mathbb{N}$, then let $ x'$ be it's "successor": $ x' \in \mathbb{N}$
[b]P3.[/b] There is no $ x$ such that $ x' \equal{} 1$.
[b]P4.[/b]  If $ x \neq 1$, then $ \exists y \in \mathbb{N}$ such that $ y' \equal{} x$.
[b]P5.[/b]  If $ S \subset \mathbb{N}$, $ 1 \in S$, and the implication $ x \in S \Rightarrow x' \in S$ holds, then $ S \equal{} N$.

We have to define $ 1' \equal{} 2$.

Then we got to have some way to add:
$ a,b \in \mathbb{N}$. If $ b \equal{} 1$ then $ a \plus{} b \equal{} a'$.(Using [b]P1, P2[/b]) Else $ \exists c \in \mathbb{N}$(Using [b]P4[/b]) such that $ c' \equal{} b$. Then $ a \plus{} b \equal{} a \plus{} c' \equal{} (a \plus{} c)'$.

From here it is easy:

$ 1 \plus{} 1 \equal{} 1' \equal{} 2$ :)

If we define $ 2' \equal{} 3, 3' \equal{} 4, 4' \equal{} 5, 5' \equal{} 6, 6' \equal{} 7, 7' \equal{} 8, 8' \equal{} 9, 9' \equal{} 10$ we can even more. For instance:

$ 2 \plus{} 2 \equal{} (2 \plus{} 1)' \equal{} (1 \plus{} 2)' \equal{} ((1 \plus{} 1)')' \equal{} (2')' \equal{} 3' \equal{} 4$ :)


Let $ n \in \mathbb{N}$, $ n$ odd, and $ A \in M_n(\mathbb{R})$. If $ A^2\equal{}O_n$ then prove that: $ \det (A\plus{}I_n) \ge \det (A\minus{}I_n).$

Use the Jordan normal form to assume WLOG that $ A$ is strictly upper triangular. Then $ \det(A\plus{}I)\equal{}1$ and $ \det(A\minus{}I)\equal{}(\minus{}1)^n\equal{}\minus{}1$.


Find the number of permutations $\sigma$ of the set $\{1,...,2n\}$ such that $2k$ and $2k+1$ aren't in neighbourhood ($1\le k\le n-1$).

Thank you for replies.

If 2k+1 cannot follow 2k, by inclusion-exclusion the number of permutations is
\[ \sum_{i=0}^n {n\choose i} (-1)^i (2n-i)!  \]
that gives rise to a sequence [url=http://oeis.org/A033815]A033815[/url].

If it is also forbidden for 2k to follow 2k+1, then the number of permutations is
\[ \sum_{i=0}^n {n\choose i} (-1)^i (2n-i)! 2^i \]
that gives rise to a sequence [url=http://oeis.org/A007060]A007060[/url].

I really still misunderstood something!Plz explain carefully for us,maxal!!

That is really simple as soon as one knows the [url=http://en.wikipedia.org/wiki/Inclusion-exclusion]Inclusion-Exclusion[/url] principle.

Could you explain clearly how did you use this principle :?:


function
functional equation
Inequality
IMO Shortlist
algebra

For an integer $ m$, denote by $ t(m)$ the unique number in $ \{1, 2, 3\}$ such that $ m \plus{} t(m)$ is a multiple of $ 3$. A function $ f: \mathbb{Z}\to\mathbb{Z}$ satisfies $ f( \minus{} 1) \equal{} 0$, $ f(0) \equal{} 1$, $ f(1) \equal{} \minus{} 1$ and $ f\left(2^{n} \plus{} m\right) \equal{} f\left(2^n \minus{} t(m)\right) \minus{} f(m)$ for all integers $ m$, $ n\ge 0$ with $ 2^n > m$. Prove that $ f(3p)\ge 0$ holds for all integers $ p\ge 0$.

[i]Proposed by Gerhard Woeginger, Austria[/i]

[hide="some observations"](will try to finish later) We claim that $ f(2k - 1) + f(2k) + f(2k + 1) = 0\quad (*)$ by strong induction, obviously true for base case. Indeed, suppose that $ 2k \neq 2^n$ for any integer $ n$, and let $ 2k - 1 = 2^n + o$ where $ 2^n > o$ and $ o$ is odd. Then
\begin{align*}f(2k - 1) + f(2k) + f(2k + 1) & = \left[\sum_{i \in \{1,2,3\}} f(2^n - i) \right] - \left[f(o) + f(o + 1) + f(o + 2)\right] \\
& = 0.\end{align*} On the other hand, if $ 2k = 2^n$, then $ f(2^n - 1) + f(2^n) + f(2^n + 1) = f(2^n - 1) + f(2^n - 3) + f(2^n - 2) - f(0) - f(1) = 0$ as desired.

For lack of time, the next couple of claims will lack rigor. The claim is that $ f(6k) - f(6k - 3) = 1$ and $ f(6k + 2) - f(6k - 1) = - 1$ [i]most[/i] of the time (it may be $ - 1$ and $ 1$ respectively sometimes, when $ l < 2^n < l + 3$ for some $ n$, or if an issue carries over from $ f(l - 2^n)$). Both, with $ (*)$, would give that $ f(3k) - f(3k - 3) = 1$, occasionally equal to $ - 1$. Now consider $ f(k) - f(k - 3)$; using  $ 2^n < k - 3 < k < 2^{n + 1}$, recursing twice gives (usually) would give, because $ f(2^n - t(m))$ will (usually) be the same, $ f(k) - f(k - 3) = f(k_2) - f(k_2 - 3)$ for some $ k_2 < 2^{n - 1}$. Now writing out the terms of the sequence from $ f( - 1), f(0), \ldots, f(32)$ gives $ 8$ of $ 32 + 3$ 'erronous' results, namely $ k = 2, 7, 14, 17, 18, 20, 26, 31$. From this it seems one could build an induction+casework argument without too much difficulty.
[/hide]

I've found a solution, but it's really messy.  I had to prove about 8 conjectures by collective strong induction.  Is there a better intended solution?

[quote="MellowMelon"]Not sure what this problem was doing on the shortlist.[/quote]

It's probably there because it probably was one of the seven best original algebra problems in the long list.

As for a solution, the best that I can find is using induction to find $ f(2^n \minus{} a)$ for all $ n$ and $ a \equal{} 1,2,3$ (you get a different -simple- expression whether $ n$ is even or odd), then find a bound for $ |f(m)|$, unless I am mistaken you get
[hide]$ |f(m)|\leq3^\frac {k}{2}$ when $ m < 2^k$[/hide]
and then write for any $ 2^n\leq r < 2^{n \plus{} 1}$
[hide]$ r \equal{} 2^n \plus{} 2^{n'} \plus{} m$ with $ n' < n$ and $ 0\leq m < 2^{n'}$, leading to $ f(r) \equal{} f(2^n \minus{} t(2^{n'} \plus{} m)) \minus{} f(2^{n'} \minus{} t(m)) \plus{} f(m)$, where $ r$ is a multiple of $ 3$ iff $ 2^n \minus{} t(2^{n'} \plus{} m)$ is a multiple of $ 3$, and if $ r$ is a multiple of $ 3$, then $ 2^{n'} \plus{} m$ is not and neither is $ 2^{n'}\minus{}t(m)$.  The first term is at least $ 3^\frac {n \minus{} 1}{2}$, the second term is non-negative since it is substracting a non-positive number, and the third term is in absolute value no more than $ 3^\frac {n'}{2}\leq3^\frac {n \minus{} 1}{2}$.[/hide]

An ugly approach(very sorry if repeated):

Use induction to show:
Case 1:if $ \minus{}1\le x \le 2^{2m}\minus{}1$ then the maximal value is $ f(2^{2m}\minus{}1)$ the minimal value is $ f(2^{2m}\minus{}2)$
Case 2:if $ \minus{}1\le x \le 2^{2m\plus{}1}$ then the maximal value is $ f(2^{2m\plus{}1}\minus{}2)$ the minimal value is $ f(2^{2m\plus{}1}\minus{}1)$

Then it is easy to show $ f(x)\le 0$ when $ x$ is not a multiple of $ 3$ and $ f(3p)\ge 0,p\ge 0$

Very sorry for not being completed,the induction is not difficult but rather complicated :P

[quote="MellowMelon"][hide="What you appear to be headed towards"]$ f(n\plus{}3)\minus{}f(n) \equal{} (\minus{}1)^b$, where $ b$ is the number of $ 1$s in the binary expansion of $ n$ (or maybe binary expansion of $ n\plus{}3$, can't remember).[/hide]

It seems we should be able to finish from this, but I was not able to do so after a lot of trying. Looking at completed solutions gives reasons to suspect it is actually very difficult to complete from here (as in, you may as well do something else). Very disappointing. :([/quote]

I have a solution based on that, not sure if it works. The last inequality is probably off a little but I think it can be easily fixed using what I proved.

Solved with [b]squareman, megarnie, vsamc, pi271828, notsarcastic69, fuzimiao2013[/b].

Omitted some trivial computational details which take a long time to type.
--------
First, list out $f(n)$ in $-1 \le n \le 8$, and check that they satisfy the given requirement. This takes care of edge cases.

[color=#f00][b]Claim:[/b][/color] For all $m \ge 2$, it holds that $f(3m) \ge 2$, $f(3m+1) \le -1$, and $f(3m+2) \le -1$.

[color=#00f][b]Proof:[/b][/color] Proceed by triple induction, assuming that for $6 \le n \le 2^{2k-1} + 2$, $f(n)$ satisfies the above requirements depending on its modulo 3. The reason we use $2k-1$ instead of $k$ is because then, its modulo 3 is fixed. We will prove that it holds up to $2^{2k+1}+2$ as well.

Define $g(n) = f(n+3) - f(n)$. We can easily deduce that $$\left(g(2^{2k-1}), g(2^{2k-1} + 3), g(2^{2k-1}+6), \cdots g(2^{2k}-5), g(2^{2k}-2) \right)= \left( -g(0), -g(3), -g(6), \cdots -g(2^{2k-1}-2), +1 \right)$$
$$\left(g(2^{2k-1}+1), g(2^{2k-1} + 4), g(2^{2k-1}+7), \cdots g(2^{2k}-4), g(2^{2k}-1) \right)= \left( -g(1), -g(4), -g(7), \cdots -g(2^{2k-1}-4), +1 \right)$$
$$\left(g(2^{2k-1}+2), g(2^{2k-1} + 5), g(2^{2k-1}+8), \cdots g(2^{2k}-6), g(2^{2k}-3) \right)= \left( -g(2), -g(5), -g(8), \cdots -g(2^{2k-1}-6), -1 \right).$$

Specifically, the three sequences are 2, 0, 1 modulo 3 respectively. Using the fact that sums of $g$'s telescope, and the values of $f(0), f(1), f(2)$, we can easily verify that the claimed inequalities hold up to $2^{2k}+2$. Then, we apply the process again (although slightly different because modulo 3's change, but the idea is the same).

$$\left(g(2^{2k}+1), g(2^{2k} + 4), g(2^{2k}+7), \cdots g(2^{2k+1}-6), g(2^{2k+1}-3) \right)= \left( -g(1), -g(4), -g(7), \cdots -g(2^{2k}-6), -1 \right)$$
$$\left(g(2^{2k}+2), g(2^{2k} + 5), g(2^{2k}+8), \cdots g(2^{2k+1}-5), g(2^{2k+1}-2) \right)= \left( -g(2), -g(5), -g(8), \cdots -g(2^{2k}-5), +1 \right)$$
$$\left(g(2^{2k}), g(2^{2k} + 3), g(2^{2k}+6), \cdots g(2^{2k+1}-4), g(2^{2k+1}-1) \right)= \left( -g(0), -g(3), -g(6), \cdots -g(2^{2k}-4), +1 \right).$$

Using telescoping, we can once again easily show that the inequalities hold for each of 2, 0, 1 modulo 3, completing the inductive step.

Specifically, we have that $ f(3p)\ge 0$ for all all integers $ p\ge 0$.$\blacksquare$

(I might've messed up indexing of the modulos a bit but thats the main idea)

[quote=MellowMelon]That's what the official solution pretty much is, and I have yet to see anything better.

Not sure what this problem was doing on the shortlist.[/quote]

To add, why is it not NT? A very lame problem.

Brutal problem. Here is a proof written backwards with motivation (to write it forwards, just reverse the paragraphs):
$$3p=2^a+2^b+c\text{ }\forall a>b \text{ and } 2^b>c\ge 0\implies f(3p)=f(2^a+2^b+c)=f(2^a-t(2^b+c))-f(2^b-t(c))+f(c).$$ So based off of this, we want to find an upper bound for f(divisible by 3 term) (since $2^n+m\equiv 2^n-t(m)\equiv 2^n+m\pmod 3$), an upper bound for f(some term not divisible by 3), and an upper bound for general $c<2^b$.

Let's compute the general values of the LHS of original statement, taking m in {1,2,3}. (since it's similar to t(m) and then we can use inductive hypothesis more). We will prove that $f(2^{2k+1}-\{1,2,3\})=\{-3^k,0,3^k\}$ and $f(2^{2k+2}-\{1,2,3\})=\{3^k2,-3^k,-3^k\}$. Check the base cases manually (I don't wanna type them rn). Now assume these statements are true for all values $l<m$. Inductive step: $$f(2^{2k+1})=f(2^{2k}+(2^{2k}-1)=f(2^{2k}-3)-f(2^{2k}-1)=-3^{k-1}-3^{k-1}2=-3^k.$$ Similarly we can derive the other ones, which I'll omit here. Then the key takeaways are $$|f(2^n-t(m)|\le \dfrac23 3^{n/2},\text{ }f(2^n-t(m))\ge 3^{(n-1)/2}\text{ }\forall 2^n+m\equiv 0\pmod 3, \text{ and } f(2^n-t(m)\le 0\forall 2^n+m\ne 0\pmod 3.$$ These are our upper bounds for divisible and not divisible by 3 inputs for f. $\square$

Next, we prove by induction that $f(m)\le 3^{n/2}\text{ }\forall 2^n>m,n\ge 0$. Base case clear as f(0)=1, and inductive step says if $m<2^n$ we're done, $$m=2^n+q\text{ }\forall 2^n>q\ge 0\implies |f(m)|=|f(2^n-t(q))-f(q)|\le |f(2^n-t(q)+|f(q)|\le \dfrac23 3^{n/2}+3^{n/2}<3^{(n+1)/2},$$ as claimed. $\square$
Putting it all together, $f(c)\ge -3^{b/2}$, $f(2^a-t(2^b+c)\ge 3^{(a-1)/2}$, $f(2^b-t(c))\le 0$ $$\implies f(3p)\ge 3^{(a-1)/2}-3^{b/2}\ge 0\text{ }\forall a>b,$$ as desired. $\blacksquare$


algebra
polynomial
geometry
symmetry
quadratics
USAMTS
Rational Root Theorem

I was thinking that perhaps we could establish a new thread once in a while about a general area of problem-solving and then we could all compile a list of theorems/tricks/techniques that are commonly used for that topic. I just randomly picked the topic of polynomials to start off with. Here are some of the techniques that I can think of off the top of my head right now. Please add your own, and ask questions if you don't understand a technique that someone names

- Descartes' Rule of Signs
- Rational Root Theorem
- Viete's Formulas
- Exploiting symmetry (subsituting y=x+1/x)
- Transforming polynomials
- Newton's Sums (Newton-Girard identities)
- Roots of Unity
- When given something like (x-a)(x-b)(x-c)(x-d), try to make a substitution that would eliminate the x terms and reduce the polynomial to a quadratic

..ok, so maybe I don't know very many tricks.  :D

One very simple thing that is often helpful is that a polynomial only has a finite number of roots..

If you are given that a polynomial is of degree n, and you are able to prove it has n+1 roots, that polynomial must be the constant zero polynomial.

Remember, if you have all the roots of the polynomial, you know its factorization (except for the leading term).

dont forget that irrational and complex roots come in conjugate pairs, and thus a polynomial with odd degree has to have at least one real root.

how would you use the roots of unity for a polynomial prob?

[quote]how would you use the roots of unity for a polynomial prob?[/quote]

Can't think of anything right now, but they come up every once in a while on both polynomial and non-polynomial problems.

Yeah roots of unity come up all the time. I never [i]really[/i] understood exactly why it works, but for example you can prove
x^4 + x^3 + x^2 + x + 1 is a factor of x^28 + x^21 + x^14 + x^7 + 1 because when you substitute the 5th roots of unity into the second part there, you get x^3 + x + x^4 + x^2 + 1, which is the same as the factor, so somehow that proves its a factor. Something to do with them both having the same roots..

sounds interesting. i will try doing something like this. yay 100th post.

That thing with a poly of deg n and n+1 roots works in every field, so we could take Zp where p is a prime:

If p|P(xi) with i from 1 to n+1 and P has deg n and xi are all different modulo p then p divides all the coefficients of the polynomial P.

This is quite useful. If I can remember correctly Wilson's thm has a proof based on this. It has, of course, other types of solns, but this works too. You just have to find the right polynomial.

one thing i found to be valuable was approximation (when you are working with [b]integers[/b]). like for instance if you have an equation like n(n+1)=4000 (from the mc mathjam) and you know your answer will be an integer, just take the average (n+0.5), then it becomes (n+0.5) :^2:, so then take square roots and your answer will be very close to the actual one. [/b]

This is useful theorem about polynomials with integer coefficients:
If $a$ and $b$ are distinct integers and $W(x)$ is a polynomial with integer coefficients then:

$a-b|W(a)-W(b)$

Seems like no one mention about just factoring...(ex: USAMTS round 1 number 3).

what are
- Exploiting symmetry (subsituting y=x+1/x)
- Transforming polynomials
- Newton's Sums (Newton-Girard identities)
- Roots of Unity
sry, new to math beyond my school textbook  :blush:

- When given something like (x-a)(x-b)(x-c)(x-d), try to make a substitution that would eliminate the x terms and reduce the polynomial to a quadratic
is this like creating a heron's scenario?  make x=a+b+c+d and plugging it in....making it 3x more messier than it already is...


integration
calculus
calculus computations

Find $ \int \frac{\sqrt{x\minus{}x^3}}{1\minus{}x^4}\,dx$.

If you need primitive then you can use mathematica by this you can save time. If you post it for student's then I agree with you. this is good examples.  :)


99 bottles of beer on the wall, 99 bottles of beer, You take one down, you pass it around, 98 bottles of beer on the wall.

I have a feeling we will fail dismally....

98 bottles of beer on the wall, 98 bottles of beer, You take one down, you pass it around, 97 bottles of beer on the wall.

97 bottles of beer on the wall, 97 bottles of beer, you take one

down, pass it around, 96 bottles of beer on the wall

-----------------------------------------------------------

Can it just be 2 players posting as long as there isn't a double-post?

96 bottles of beer on the wall, 96 bottles of beer, You take one down, you pass it around, 95 bottles of beer on the wall.

95 bottles of beer on the wall, 95 bottles of beer, You take one down, you pass it around, 94 bottles of beer on the wall.

94 bottles of beer on the wall, 95 bottles of beer, You take one down, you pass it around, 93 bottles of beer on the wall.

That's why copying and pasting can be bad sometimes. But I still do it :D
93 bottles of beer on the wall, 93 bottles of beer, You take one down, you pass it around, 92 bottles of beer on the wall.

92 bottles of beer on the wall, 92 bottles of beer, You take one down, you pass it around, 91 bottles of beer on the wall.

91 bottles of beer on the wall, 91 bottles of beer, you take

one down, pass it around, 90 bottles of beer on the wall

-----------------------------------

Klebian: You made a Typo  :o  more then once..

---------------------------------

_______

Maybe we aren't dead yet  :)

90 bottles of beer on the wall, 90 bottles of beer, You take one down, you pass it around, 89 bottles of beer on the wall.

89 bottles of beer on the wall, 89 bottles of beer, you take one

down, you pass it around, 88 bottles of beer on the wall.

-----

2 of us get together to win :)

88 bottles of beer on the wall, 88 bottles of beer, you take one

down, you pass it around, 87 bottles of beer on the wall.

87 bottles of beer on the wall, 87 bottles of beer, you take one

down, you pass it around, 86 bottles of beer on the wall.

86 bottles of beer on the wall, 86 bottles of beer.
Take one down, pass it around, 85 bottles of beer on the wall.

85 bottles of beer on the wall, 85 bottles of beer, you take one

down, you pass it around, 84 bottles of beer on the wall

_________________

I told you that we would fail dismally

84 bottles of beer on the wall, 84 bottles of beer, you take one

down, you pass it around, 83 bottles of beer on the wall.

We need like everybody online at once.

83 bottles of beer on the wall, 83 bottles of beer.
Take one down, pass it around, 82 bottles of beer on the wall.

Eh, I'm extending this another $ 10\pi$ hours. Eh.[size=9]P.S. find the loophole[/size]

[quote="miyomiyo"]Perfect said to find the loophole, so I'm trying this out.
Somebody must post all the remaining numbers in one post.  There is no violation of any rule in doing so.[/quote]

when did he say this?


Can anyone teach me how to prove these two?

Firstly, try to avoid using images. You can learn $\text{\LaTeX}$ at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/url] and write problems in a legible form.
Secondly, the first is obvious. We have to prove that for any $\varepsilon>0$ there exists such $\delta>0$ that for any $-\delta<x<\delta$ we have $|g(x)|<\epsilon$. But we can take $\delta_\epsilon= \epsilon/2$, then for $-\epsilon/2<x<\epsilon/2$ we have $|g(x)|=\begin{cases}{|x|\text{ for }x\text{ rational}\\{0\text{ for }x\text{ irrational}\end{cases}}$ which is obviously less than $\varepsilon$ which ends the proof.

For the second: for $\epsilon =1$ there exist a $\delta > 0$ such that

for all  $x$ with $0<|x-c| < \delta$ we have $| f(x)-\ell | < 1$

and this is equivalent to $\ell-1 < f(x) < \ell+1$ .

Now take $B = max \{ |\ell+1|, |\ell-1| \}$ .

I am sorry that I have make the inconvenience here. I will take your advice next time.

Thank you very much for solving my problems.


can anyone plz suggest me software for learning permutations n combinations,ie a software which lets me generate permutations n see for myself the patterns etc..

almost any calculator should be able to do permutations and combinations

if you want to see patterns in combinations, just draw yourself a nice pascal's triangle

$\binom{6}{n}$ would be the row with 7 numbers.

Each of those 7 numbers would represent an $n$ from 0 to 6 in order


prove that Nagel point and exhemothecy center of incircle and circumcircle of triangle $ABC$ are isogonal conjugate.

Thank you, but I already solved this one here: http://www.mathlinks.ro/Forum/viewtopic.php?t=23668 .

  darij

Nice Darij.

there is another solution with inversion.


Does anyone know where I can find the Mathcounts State competitions from before 2001? The old links no longer work.

I can always get old tests from my coach.

I mean online.

I have a bunch of old nationals, but I m more interested in states.

I remember there was a link to one site, but it only had 2002 onwards.

But, I believe 2002 onwards will be better preparation than 2001 before.  They are harder in my opinion.

The two hardest states I've seen are 2005 and 1999.

That is true.

Oh, yeah, never mind, before 2001 States are somewhat hard as well.

I also think 1996 States was somewhat difficult as well.

I will be willing to trade any state tests from 1990 onwards in exchange for chapter tests (please be generous, maybe something like 3 chapters for every 2 states)

You're still doing MATHCOUNTS next year?

No, but I'll be assisting my school, and while I have plenty of states and nationals, what those guy swould need to practice on are chapters. oh, and school tests as well
 (btw, I will also trade state and/or national tests for school tests; these are my rates (sorry if it seems "expensive): 3 chapter for 2 states, 2 chapter for 1 nationals, 2 schools for 1 state, 3 schools for 1 national... or we could bargain  :wink: )

Er, by "2 schools", I mean 2 school tests, lest Neil come and ask how he's gonna e-mail me his school...

[quote="kheg_k"]Do you mean chapters before 2001?[/quote]

Yes, sorry

Klebian, shouldn't Nationals be worth the least because they're pretty easy to find especially with the search button? It's not Let's Make A Deal, we should just all try to help each other when we need tests.


analytic geometry
combinatorics unsolved
combinatorics

On a coordinate plane, there are three coins at points $(0,0), (0,1),(1,0)$. You can remove any coin but you must replace it with two coins, one directly to the right and another one directly above. Is it possible that there remains no coin on those three coordinates?

After a finite number of moves, the answer is no.
For each point $(x,y)$ let $\frac 1 {2^{x+y}}$ be its weight.

The initial sum of weights of the points which have coins is $1+\frac 1 2 + \frac 1 2 = 2.$
In another hand, any removal leaves the total sum unchanged because we replace $\frac 1 {2^{x+y}}$ by $2 \cdots \frac 1 {2^{x+y+1}}$.
But, if the three initial points are free of coin then the total sum is less than $\sum_{x,y \geq 0} \frac 1 {2^{x+y}}-(1+\frac 1 2 + \frac 1 2 ) = 4- 2 = 2$, hence result.

Pierre.


If $2^{x+3}+2^{x}=3^{y+2}-3^{y}$ and $x$ and $y$ are integers, find all solutions of $(x,y)$.

Please, no guess and check. :)

[quote="Totally Zealous"][quote="klutzycherry14"][quote="DanK"][hide]$2^{x+3}+2^{x}=3^{y+2}-3^{y}$
$2^{x}(2^{3}+1)=3^{y}(3^{2}-1)$
$2^{x}3^{2}=2^{3}3^{y}$
$(x,y) = (3,2)$[/hide][/quote]

[hide]umm..i was just wondering something, when u factored the equation to $2^{x}(2^{3}+1)=3^{y}(3^{2}-1)$ , if you unfactor it, would it not equal $4^{3x}+2^{x}$=$9^{2y}-3^{y}$?? i was just wondering if that is correct way to factor that equation. it is probably right, and im just missing something, because im  not that good at factoring... :blush:
 [/hide][/quote]
yes Dank is right. Bases don't change when you factor.[/quote]

...and when you distribute.  $x^{a}\cdot x^{b}=x^{a+b}$

[quote="robinhe"][/quote][quote="Totally Zealous"][quote="klutzycherry14"][quote="DanK"][hide]$2^{x+3}+2^{x}=3^{y+2}-3^{y}$
$2^{x}(2^{3}+1)=3^{y}(3^{2}-1)$
$2^{x}3^{2}=2^{3}3^{y}$
$(x,y) = (3,2)$[/hide][/quote]

[hide]umm..i was just wondering something, when u factored the equation to $2^{x}(2^{3}+1)=3^{y}(3^{2}-1)$ , if you unfactor it, would it not equal $4^{3x}+2^{x}$=$9^{2y}-3^{y}$?? i was just wondering if that is correct way to factor that equation. it is probably right, and im just missing something, because im  not that good at factoring... :blush:
 [/hide][/quote]
yes Dank is right. Bases don't change when you factor.[/quote][quote="robinhe"]

...and when you distribute.  $x^{a}\cdot x^{b}=x^{a+b}$[/quote]

oh..that's cool...didn't know that..thanks! :D


I know it's possible to go from tex to pdf, but what about pdf back to a latex source file? Is it possible? (When I open a pdf in Texnic center, it has these strange symbols).

It is difficult to do. As [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=recovertex]TeX FAQ[/url] says:

[quote]The job just can't be done automatically: DVI, PostScript and PDF are "final" formats, supposedly not susceptible to further editing - information about where things came from has been discarded.[/quote]However, it does go on to make some suggestions.

You'll find further discussion of this problem on [url=http://groups.google.co.uk/groups?hl=en&lr=&c2coff=1&q=%22pdf+to+tex%22]Google Groups[/url]


limit
inequalities
real analysis
real analysis unsolved

$\left( {u_n } \right)$ is a sequence of positive reals such that $\mathop {\lim }\limits_{n \to + \infty } \left( {u_n + \dfrac{1} {2}u_{2n} } \right) = 1$
find $\mathop {\lim }\limits_{n \to \infty } u_n$

If the limit exists, then it can only be $\dfrac {2}{3}$, for if $u_{n} \rightarrow x$, then $x+\dfrac {1}{2}x=1$. Now you have to prove the existence.

Your sequence $(u_{n})$ is bounded (otherwise -and by positivity of $u_n$-) $\forall N, \exists K \geq N, u_{K} \geq 100$.) But since for such a $K$ sufficiently large we must also have $u_{K}+\dfrac {1}{2}u_{2K}$ close to $1$, that means $u_{2K}$ will be close to $2-2u_{K} < 0$, which is absurd since $u_n \geq 0$. (I leave it to you to write everything formally)

Now you know $u_{n}$ is bounded write $v_{n}=u_{n}-\dfrac {2}{3}$. If we can prove that $lim \ v_{n}=0$ then we are done. But given $\epsilon>0$ and for $n$ sufficienly large,

$|\dfrac {(-1)^{k}}{2^{k}}v_{2^{k}n}+\dfrac {(-1)^{k}}{2^{k+1}}v_{2^{k}n}| \leq {\dfrac {1}{{2^{k}}}} \epsilon$.

Now adding those inequalities and considering that $u_{n}$ is bounded proves that $lim \ v_{n}=0$


$ x,y,z > 0$ . Prove that !
$ \sum \frac {1}{(x \plus{} y)^2} \ge \frac{9}{4(xy \plus{} yz \plus{} xz)}$

[quote="bigbang195"]$ x,y,z > 0$ . Prove that !
$ \sum \frac {1}{(x \plus{} y)^2} \ge \frac {9}{4(xy \plus{} yz \plus{} xz)}$[/quote]

$ \Longleftrightarrow \sum_{sym} (4x^5y \minus{} x^4y^2 \plus{} x^4yz \minus{} 3x^3y^3 \minus{} 2x^3y^2z \plus{} x^2y^2z^2)\geq 0$.

 true due to AM-GM Inequality and Muirhead theorem .

or

$ \Longleftrightarrow 4\sum x^3(y \plus{} z)(x \minus{} y)(x \minus{} z) \plus{} 3(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \plus{}$ $ 16xyz\sum x(x \minus{} y)(x \minus{} z) \plus{} 4xyz\sum (y \plus{} z)(x \minus{} y)(x \minus{} z)\geq 0$.

[quote="bigbang195"]$ x,y,z > 0$ . Prove that !
$ \sum \frac {1}{(x \plus{} y)^2} \ge \frac {9}{4(xy \plus{} yz \plus{} xz)}$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3547


algorithm
articles
modular arithmetic
function
number theory
Euclidean algorithm

What are the methods used to solve simultaneous congruences?  I don't think CRT would help very much...

That's precisely what CRT is for.  Could you be more specific?

Well, CRT states that $ x_1 \equiv x_2 \mod a_1a_2a_3 \cdots a_n$ where $ x_1$ and $ x_2$ are particular solutions to
$ x \equiv b_k \mod a_k (1\le k\le n)$, but how does it help to find any particular solution in the first place?

That's the less specific version of CRT.  Read about the more specific version, for example, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]here[/url].  See also the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem#A_constructive_algorithm_to_find_the_solution]Wikipedia article[/url].

Could you please give me an example by solving for $ x$ in $ x\equiv 0\mod 3\equiv 4 \mod 5$?  Thanks.

Certainly.  To solve $ u \equiv 0 \bmod 3$ and $ u \equiv 1 \bmod 5$ we want to find the multiplicative inverse of $ 3 \bmod 5$, which we do by the extended Euclidean algorithm (or by inspection); it's $ 2$.  Then $ u \equiv 6 \bmod 15$, hence

$ x \equiv 4u \equiv 24 \equiv 9 \bmod 15$.

(If $ x$ weren't divisible by $ 3$ we would also have had to find $ v$ such that $ v \equiv 1 \bmod 3, v \equiv 0 \bmod 5$.)  Generally, the solution to $ x \equiv a \bmod 3, x \equiv b \bmod 5$ is

$ x \equiv bu \plus{} av \bmod 15$.

@mathwizarddude:
To solve your given example, you could also use what is called the "method of successive subsitution" (for obvious reasons).
$ x\equiv 0\pmod 3\implies x\equal{}3m$ for some $ m\in\mathbb{Z}$ (this notation means that $ m$ is a member of the set of integers).

Substituting into the second congruence:
$ 3m\equiv 4\pmod 5$. We must now find the modular multiplicative inverse of $ 3\bmod 5$, which is $ 2$.

Therefore, $ 2\cdot 3m\equiv m\equiv 2\cdot 4\equal{}8\equiv 3\pmod 5$.
$ m\equiv 3\pmod 5\implies m\equal{}5n\plus{}3$ for $ n\in\mathbb{Z}$.

Substituting back, we get that $ x\equal{}3(5n\plus{}3)\equal{}\boxed{15n\plus{}9}$.

Try it; any integer $ n$ can be plugged into that function to give a value of $ x$ that fits both congruences. Indeed, you can show that this holds, because $ 15n\plus{}9\equiv 0\pmod 3$ and $ 15n\plus{}9\equiv 9\equiv 4\pmod 5$.


Could someone help me find out how to make 3 by 3 grids in LaTeX for Round 3 Problem 2? It's not really helping me out with the solution, technically; it's just helping with LaTeX formatting.(I started last week) I have no idea how to do things other than limits, fractions, and whatever I can copy and paste?

You can go to google and type in "latex tables".. (this question would be better asked in teh latex forum)

It doesn't exactly make a grid, but it should work for the situation.

$ \begin{pmatrix}
a_1 & a_2 & a_3\\
b_1 & b_2 & b_3\\
c_1 & c_2 & c_3\\
\end{pmatrix}$

[code]$\begin{pmatrix}
a_1 & a_2 & a_3\\
b_1 & b_2 & b_3\\
c_1 & c_2 & c_3\\
\end{pmatrix}$[/code]

Not quite the same, but closer.
$ \begin{tabular}[t]{|c|c|c|}
\hline1&2&3\\\hline
4&5&6\\\hline
7&8&9\\\hline
\end{tabular}$
[code]\begin{tabular}[t]{|c|c|c|}
1&2&3\\\hline
4&5&6\\\hline
7&8&9\\\hline
\end{tabular}[/code]

I made this in jPicEdt, if you install it, make sure you read the FAQ... (also make sure you have Java on your computer)

[code]\begin{picture}(130,90)(0,0)
\linethickness{0.3mm}
\put(40,90){\line(1,0){90}}
\put(40,0){\line(0,1){90}}
\put(130,0){\line(0,1){90}}
\put(40,0){\line(1,0){90}}
\linethickness{0.3mm}
\put(40,30){\line(1,0){90}}
\linethickness{0.3mm}
\put(40,60){\line(1,0){90}}
\linethickness{0.3mm}
\put(70,0){\line(0,1){90}}
\linethickness{0.3mm}
\put(100,0){\line(0,1){90}}
\end{picture}[/code]

Well, if you're not going to make it with Latex code, then it's pretty easy to do it with a drawing program and save it as a pdf.  Normally for geometry problems, I just use Powerpoint or Canvas to draw it.


[quote="GameBot"]The difference when the product of 4 and a number is subtracted from 16 is greater than 10. How many positive integers satisfy the given conditions?[/quote]

Let the number be $ n$. We can then write the inequality $ 16 \minus{} 4n > 10$. Subtracting 16 from both sides and then dividing by -4 yields $ \minus{}4n > \minus{}6 \implies n < 1.5$. Thus $ \boxed{1}$ positive integer satisfies the given conditions.


Let $a,b,c$ be the sides of a triangle. Prove that $\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}}\geq \frac{a}{c}(a+b-c)+\frac{c}{b}(c+a-b)+\frac{b}{a}(b+c-a)$.

Shouldn't this be posted in the inequalities forum? :lol:


inequalities
rotation
conics
ellipse
algebra
polynomial
number theory unsolved

[quote="ummagumma"]Find the minum value of a so that:

6x^2 + y^2 + a >= 4xy + y

For this value, find:
a) all the integer couples (x;y)
b) all the real couples (x;y)
that are solutions.

It's a rotated ellipse, but i can't write it as a polynomial sum.
enjoy![/quote]

The inequality may be written as $ 2(x-\frac{1}{2})^{2}+(2x-y+\frac{1}{2})^{2}+a-\frac{3}{4}\geq 0$ $ \forall x,y$

If $ (x,y)\in\mathbb{R}^{2}$, the minimum value for $ a$ is $ a=\frac{3}{4}$ and the equality is then reached for $ x=\frac{1}{2}$ and $ y=\frac{3}{2}$

If $ (x,y)\in\mathbb{N}^{2}$, the minimum value for $ a$ is $ a=0$ and the equality is reached for $ x=1$ and $ y=2$

[quote="ummagumma"]don't you consider the solution (0,0) for x,y natural?[/quote]

Although in my country the set $ \mathbb{N}$ of natural numbers contains the number 0, it seems that this is generally not the case elsewhere. So, on this forum, without explicit precision, I consider generally that 0 is not to be considered.


A box contains gold coins. If the coins are equally divided among
six people, four coins are left over. If the coins are equally
divided among five people, three coins are left over. If the box
holds the smallest number of coins that meets these two conditions,
how many coins are left when equally divided among seven people?

If there are $ 3$ coins left when divided among $ 5$ people, then the number must end in a $ 3$ or an $ 8$.

If there are $ 4$ left over when divided among $ 6$ people, then the possible numbers of coins are:

$ 4,10,16,22,28,34,40,...$

The only one that ends with $ 3$ or $ 8$ is $ 28$, so there are $ 28$ coins, and there are $ 0$ left over when divided among $ 7$ people.


Steve has a isosceles triangle with base 8 inches and height 10 inches. He wants to cut it into 8 pieces having equal area by cutting from a point on the base of the triangle to the top vertice. To the nearest hundredth of a inch, what is the number of inches in the greatest perimeter among the 8 peices?

Even though I can't solve it, please hide your solutions.

Billy

[hide]Lets try this:
The one with most perimeter is obvoiusly the one farthest away from the center.
The triangle is total 40 (area)
So smaller must have 5 (area)
10x/2=5 -->  5x=5 --> x=1
Now use pythag theorum for other two sides. (since h and b are perp)
rt(10^2+3^2)=rt(109)
rt(10^2+4^2)=rt(116)=2rt(29)
>>1+rt(109)+2rt(29)<<[/hide]
I won't compute for laziness

[hide]If the triangle goes to the top vertice of the larger triangle, it must have a base that is 1/8 of the base of the larger triangle, so its base is 1. The triangle that is farthest from the center must be the one with the most perimeter. If the height is drawn from the top vertice to the bottom, we can form right triangles to find all of the dimensions. The base is of course 1. The right side is the square root of 9 + 100 and the left side is the square root of 116 which simplifies to 2 * sqrt29. The answer should be sqrt109 + 1 + 2*sqrt29.[/hide]


[hide="This should get you going..."]Square the second equation to get $ 24232^2 \equal{} (xy)^2(x^4\plus{}y^4\plus{}2(xy)^2) \equal{} (xy)^2(32657 \plus{} 2(xy)^2) \equal{} 2t^2 \plus{} 32657t$, where $ t \equal{} (xy)^2$. This is a quadratic in $ t$...[/hide]

[quote="luiseduardo"]x^4 + y^4=32657

xy.(x²+y²)=24232

How to do ?[/quote]
solution :
(x,y)=(+-8,+-3)=(+-3,+-8)
almonkez

[quote="luiseduardo"]x^4 + y^4=32657

xy.(x²+y²)=24232

How to do ?[/quote]
 :play_ball:

hi.  i'm currently thinking if a certain trig. substitution is possible here...but can't figure out how, if any?

what do you think guys ?   :maybe:

hello, yes the answer is
$ x \equal{} \minus{} 13,y \equal{} \minus{} 8$
$ x \equal{} 13,y \equal{} 8$
$ x \equal{} 8,y \equal{} 13$
$ x \equal{} \minus{} 8,y \equal{} \minus{} 13$
Sonnhard.

[quote="luiseduardo"]almonkez, is the answer (x,y)=(+-8,+-3)=(+-3,+-8)  or x = 8 and y = 13 ?[/quote]

im sorry for this mistake
i solved it right bur wrote it wrong
booth of x & y  negative or positive together
thanks

[quote="luiseduardo"]almonkez,

Just out of curiosity. Did you do the edges of the sheet ? Made using a pen ?
It's a beautiful paper.[/quote]
thanks : luiseduardo
i made it using   microsoft word 2003
 i made some papers for u in the file
with my bestwishes
almonkez

Thanks almonkez. It's a beautiful paper.


geometry
circumcircle
perpendicular bisector
national olympiad

Let C a circle center O and P a point in the plane, M is a point in C, and MN a diameter, Find the locus of the center of the triangles MNP(as M varies)

Let P' be the intersection of PO with the circumcircle of PMN, then $OP \cdot OP' = OM \cdot ON = R^{2}= \text{Constant}$. Since P' is always on the fixed line OP, so P' is also fixed. Let the circumcentre of triangle PMN be O', then O'P=O'P', therefore O' lies on the perpendicular bisector of PP'. Since PP' is fixed, so we get the locus.

PS You posted the problem at the wrong section.


find all the functions $ f: N\rightarrow N$ such that
$ f(m\plus{}f(n))\equal{}f(f(m))\plus{}f(n) (\forall m,n\in N )$

shouldn't it be $ f: \mathbb{N}_0\to\mathbb{N}_0$? :maybe:
($ 0\in\mathbb{N}_0$)
in this case it's the third problem of IMO 1996

[quote="toanIneq"]find all the functions $ f: N\rightarrow N$ such that
$ f(m + f(n)) = f(f(m)) + f(n) (\forall m,n\in N )$[/quote]

$ f(0+f(0))=f(f(0))+f(0)$ gives $ f(0)=0$ so $ \forall m\N: \ fof(m)=f(m)$
$ \forall f(x),f(y)\in Imf: \ f(x)+f(y)=fof(x)+f(y)=f(f(x)+f(y))\in Imf$
if $ Imf=\{0\}$ then $ f\equiv 0$
if $ Imf\neq \{0\}$ then $ \exists p,h\in\ N^*,\forall x\ge h: xp\in Imf\ and\ Imf\subset pN$
$ \forall x\in\mathbb{N}: \ (x+p)h=f((x+h)p)=f(xp+f(hp))=f(xp)+f(hp)=f(xp)+hp$ than $ f(xp)=xp$
it gives $ Imf=pN$ then $ f(kp+r)=kp+f(r)$ for $ r=0,1,2,..,p-1$
then we choose $ a_1,..,a_{p-1},p\in N$ arbaitrary and $ a_0=0$ $ \forall k\in N;\forall r\in\{0,1,..,p-1\}: \ f(kp+r)=p(a_r+k)$

cheking $ m=k'p+r',n=kp+r$
$ f(m+f(n))=f(k'p+r'+f(kp+r))=f((k'+k+a_r)p+r')=p(k+k'+a_r+a_{r'})$
$ f(f(m))+f(n)=f(f(k'p+r))+f(kp+r)=f(p(k'+a_{r'}))+p(k+a_r)=p(k'+k+a_{r'}+a_r)$

for exemple:
$ p=2,a_1=1$ gives $ f(n)=2[\frac{n}{2}]$
$ p=0$ gives $ f(n)=0$
$ p=1$ gives $ f(n)=n$

$ 2[\frac{m+2[\frac{n}{2}]}{2}]=2[\frac{m}{2}]+2[\frac{n}{2}]$


advanced fields
advanced fields unsolved

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87628]Canada MO 1969, #6.[/url]
EDIT : ok, I see you [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=157407]double-posted it..[/url]


logarithms
real analysis
real analysis unsolved

are the followings true? :D

the {$n!$} is densite.(over the reals $\le1$) and also:
{$log p$} is densite.
({} is the fraqtional part.for example${2.5}=.5$.)

The first question is unclear: $n!$ is always an integer, isn't it? Probably something is missing; if so, please, edit. I assume that the second question is about density of $\{\log p\}$ where $p$ runs over all primes. Then the answer is "Yes". It is enough to check that the sequence $\log p$ increases to $+\infty$, which is clear, and the difference between two subsequent elements of the sequence tends to $0$, which immediately follows from the prime number theorem.

ya fedja i had the same question there $n$ is a natural number.
so what can we say?

as result does $n!$ begins with any number? :D


A solar system contains four planets and a sun as shown.  Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; and planet D completes an orbit in 48 months.  If the four planets lie along a straight line in May of the year 2000, what is the next year that they will again lie along the same line?  Assume that the orbits are circular and that a year has exactly 12 months.

Please post your solutions in hide tags.

This is like the third time this has been posted  ;) .

[hide="And the answer is..."]The lcm of the three is 720. But since they line up when one planet is opposite the sun from the rest you half that. The number is 360 which is the year 2030[/hide]

[quote="kyyuanmathcount"]We'll give it another go, that was months ago.

Not locked  :D .[/quote]
Months ago?
Maybe one month ago, max...

Edit:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2030&t=74701
EXACTLY one month ago.  :P


Consider 6 positive numbers $ a,b,c,d,e,f$ satisfy: $ a \plus{} c \plus{} e \equal{} b \plus{} d \plus{} f$.
Prove that: $ \frac {af \plus{} be}{(a \plus{} b)(e \plus{} f)} \plus{} \frac {ad \plus{} bc}{(a \plus{} b)(c \plus{} d)} \plus{} \frac {cf \plus{} de}{(c \plus{} d)(e \plus{} f)} \geq \frac {3}{2}$

Possibly an attempt to cheat, please don't answer and lock it.

[edited  by PCO because possibly an attempt to cheat]

Sorry
We have a condition: $ a\plus{}b, c\plus{}d, e\plus{}f$ are the lenghts of a triangle!


Let two equations:  $ 4^x \minus{} 4^{ \minus{} x} \equal{} 2cosax$        $ (1)$ ; $ 4^x \minus{} 4^{ \minus{} x} \equal{} 2cosax \plus{} 4$  $ (2)$
              we have a value of $ a \equal{} a_0$ such that equation $ (1)$ has 2008 root.
          Determine the number of the roots of equation  $ (2)$ with $ a \equal{} a_0$.


function
induction
algebra solved
algebra

2)Are there any functions f:N-->N so that:
f(f(n))+f(n)=2n+3???

Solution:

Lets denote for a starter f(0)=a. We put n=0 in
f(f(n))+f(n)=2n+3 [1] and we get that f(a)=3-a we put n=a in [1] and we get that f(f(a))+f(a)=2a+3 and we get that f(3-a)=3a.
lets put n=3-a in [1] and we get that f(3a)=9-5a.

So 9-5a>=0 => a<=9/5 so we have either a=0 or a=1. but we can't have f(0)=a=0 because we would get that f(f(0))+f(0)=0, not 3. so we must have f(0)=1.

now it's all induction: prove that f(n)=n+1, for any n natural...

cheers! :D :D

Well, the question is "are there any functions.." so I think it's enough to show that f(n)=n+1 works?  :D

well, that's only part one of the problem

part two actually sais that: prove that there only exists one funtion that verifies that!!

so I guess Lagrangia' s solution solved both parts!!

thank you very much Lagrangia! bye bye!

well.. I thought a full solution may come in hand!! :D :D :D

cheers!


MIT
college
vector
ratio
inequalities
linear algebra
arithmetic sequence

Hi, I am a new student of Linear Algebra. I'm a self learner doing the mit ocw course.
Here is a question from the book Linear Algebra and Its applications:

Which of the following are subspaces of $ R^{inf}$

(a) All sequences like (1,0,1,0,...) which include infinitely many zeros.
- I think this is a subspace since it looks to be closed under addition and multiplication.
(1,0,1,0,...) + (2,0,2,0,...) = (3,0,3,0,...) - still a sequence
c(1,0,1,0,...) - still a sequence
Help on making this look like real math is appreciated  :D

(b) All sequences (x1,x2,...) with $ x_{j}\equal{}0$ from some point onward.
Since the last number always will be zero I think this is a subspace as well.
Addition will give a sequence with the last number(s) zero, and so will multiplication...

(c) All decreasing sequences: $ x_{j\plus{}1} \leq x_{j}$ for each $ j$
I see this as being a subspace as well since both addition and multiplication still gives a decreasing sequence..

(d) All convergent sequences: the $ x_{j}$ have a limit as $ j \rightarrow \inf$
Cannot compute

(e) All arithmetic progressions: $ x_{j\plus{}1}$ - $ x_{j}$ is the same for all $ j$
Addition of two arithmetic progression gives an arithmetic progression. Multiplication of a arithmetic progression by zero, does not give an arithmetic progression.
I dare say that this is not a subspace..

(f) All geometric progressions ($ x_{1}, kx_{1}, k^{2}x_{1},...$) allowing all k and $ x_{1}$
Cannot compute

Thank you very much for your time and patience!

The other fact you need to take into account is the presence of the zero-vector.  If there is no zero-vector in the set you're working with, it's not a subspace.

The ones I haven't commented on are correct, although you should actually go through the subspace criteria.  (What are they?)

[quote="Mollier"](a) All sequences like (1,0,1,0,...) which include infinitely many zeros.
- I think this is a subspace since it looks to be closed under addition and multiplication.[/quote]
What is $ (1, 0, 1, 0, ...) \plus{} (0, 1, 0, 1, ...)$?

[quote="Mollier"](c) All decreasing sequences: $ x_{j \plus{} 1} \leq x_{j}$ for each $ j$
I see this as being a subspace as well since both addition and multiplication still gives a decreasing sequence..[/quote]
What happens when you multiply by $ \minus{}1$?

[quote="Mollier"](d) All convergent sequences: the $ x_{j}$ have a limit as $ j \rightarrow \inf$
Cannot compute[/quote]
What do you know about sums and scalar multiples of convergent sequences?

[quote="Mollier"]Multiplication of a arithmetic progression by zero, does not give an arithmetic progression.[/quote]
Sure it is; it's an arithmetic progression with common difference $ 0$.

[quote="Mollier"](f) All geometric progressions ($ x_{1}, kx_{1}, k^{2}x_{1},...$) allowing all k and $ x_{1}$
Cannot compute[/quote]
What is the sum of two geometric progressions with different common ratios?

I should have a bit more systematic approach :)

(a)
Addition: No, because (1,0,1,0,...) + (0,1,0,1,...) = (1,1,1,1,...), just as t0rajir0u wrote.
Multiplication: Yes
zero-vector: I think not..or can (0,0,0,0,...) be seen as a sequence like (1,0,1,0,...)  :oops:

(c)
Addition: Yes
Multiplication: No, if multiplied by -c it gives a increasing sequence.
Zero vector: No, (0,0,0,...) is not a decreasing sequence.

(d)
Addition: Sum of two convergent sequences $ [a_{j}]_{j\equal{}1}^{inf}$ and $ [b_{j}]_{j\equal{}1}^{inf}$ converge to a+b.
It's been many years since I last dealt with sequences of any sort, so google helped me out on this one :)
Multiplication: Yes
Zero vector: I guess so, converges to zero?

(e)
Addition: Yes
Multiplication: Yes
Zero Vector: Yes

(f)
Addition: Yes, the common difference is summed.
Multiplication: Yes
zero vector: Yes, geometric progression with common difference 0.

Thank you both for replying!

[quote="Mollier"]zero-vector: I think not..or can (0,0,0,0,...) be seen as a sequence like (1,0,1,0,...)  :oops: [/quote]
Ignore the example (it seems to have seriously misled you); all we're interested in is the sequences containing infinitely many zeroes.

[quote="Mollier"]Zero vector: No, (0,0,0,...) is not a decreasing sequence.[/quote]
Yes it is.  "Decreasing" is not the same thing as "strictly decreasing"; mathematicians use the latter to denote the case that the inequality is strict.

[quote="Mollier"](f)
Addition: Yes, the common difference is summed.[/quote]
Think about this more carefully.  Suppose I gave you two geometric sequences, say $ (2^n)$ and $ (3^n)$.  The sum of these two sequences is the sequence $ (2^n \plus{} 3^n)$.  Is this a geometric sequence?

[quote]Ignore the example (it seems to have seriously misled you); all we're interested in is the sequences containing infinitely many zeroes. [/quote]

If it is the sequences containing infinitely many zeroes they are talking about, then there is a zero vector...

[quote]Yes it is. "Decreasing" is not the same thing as "strictly decreasing"; mathematicians use the latter to denote the case that the inequality is strict. [/quote]

May be a while before I can call my self a mathematician  :P So (0,0,0,...) may be looked at as decreasing, but not strictly? Funny...

[quote]Think about this more carefully. Suppose I gave you two geometric sequences, say  and . The sum of these two sequences is the sequence . Is this a geometric sequence?[/quote]

Well, for n=1 the answer is 5 which is not a geometric sequence...

Again, thank you very much!

[quote="Mollier"]So (0,0,0,...) may be looked at as decreasing, but not strictly? Funny...[/quote]
Yep.  In fact, we can say more:  the sequences that are both decreasing and increasing are precisely the constant sequences.

[quote="Mollier"]Well, for n=1 the answer is 5 which is not a geometric sequence...[/quote]
That's not the way I'd put it, but the sum of two geometric sequences is indeed not a geometric sequence.  $ 2^n \plus{} 3^n$ is not described by its initial term and a single common ratio.  (Interestingly enough, it [i]is[/i] described by the rule $ a_{n\plus{}2} \equal{} 5a_{n\plus{}1} \minus{} 6a_n$, but that's another story.)

[quote]Yep. In fact, we can say more: the sequences that are both decreasing and increasing are precisely the constant sequences.
[/quote]

That is way cool :)

[quote]That's not the way I'd put it, but the sum of two geometric sequences is indeed not a geometric sequence. $ 2^{n} \plus{} 3^{n}$ is not described by its initial term and a single common ratio. [/quote]

Thank you!


seems like evry1 is doing aCops... im still on volume 2  :(  How long did it take you to do it?

[quote="1=2"]What comes next in the following pattern:
1, 11, 21, 1211, 111221, 312211, etc.

I know what number comes next, I just don't know how to get that if I didn't know... how would one come to that conclusion???[/quote]
13112221
I don't know hwo you would find a pro way to do it, but I read them out loud
1 11 21 1211 and then after that, I got it.

ACoPs is not very much higher than AoPs Vol 2...

Basically the number after each number describes that number.
So we start with 1.
Then we describe it as "One 1", or 11.
We describe this as "Two 1's", or 21.
"One 2; One 1" 1211.
"One 1; One 2; Two 1's". 111221.
"Three 1's; Two 2's; One 1". 312211.
"One 3; One 1; Two 2's; Two 1's". 13112221.

I wrote a Java program to generate elements in this sequence:

[code]import java.util.*;

public class NumberSequence{
	public static void main(String[] args){
		Scanner s=new Scanner(System.in);
		String str="1";
		for(;;){
			System.out.println(str);
			s.nextLine();
			str=process(str);
		}
	}
	public static String process(String s){
		String str="";
		char c=' ';
		int i=0;
		for(int j=0;j<s.length();j++){
			if(s.charAt(j)!=c){
				str+=i;
				str+=c;
				i=0;
			}
			i++;
			c=s.charAt(j);
		}
		str+=i;
		str+=c;
		return str.substring(2);
	}
}[/code]

You may notice that there are never digits greater than 3 in this sequence. Here's why:

[hide]
Let's assume that there ARE digits greater than 3 in the sequence. Consider the [b]first[/b] number in the sequence that has a digit greater than three.

The number before that cannot have any digits greater than three. Therefore, this number must have more than three digits in a row. So it must have at least four in a row.

There are two cases:

1. The four digits are at the beginning.
Then the number must start with either 1111, 2222, or 3333. But these mean, respectively, One 1 One 1, Two 2's Two 2's, and Three 3's Three 3's. This will not happen. The correct way to express these are Two 1's, Four 2's and Six 3's. So we have reached a contradiction.

2. The four digits are not at the beginning.
Then they will be preceded by at least one digit. Let us call the digit directly before the four digits b. So we can write the number as:

$ ...baaaa...$

This can mean two things. This can either mean: b a's, a a's, which is a contradiction, or a a's, a a's, which is also a contradiction.

Thus, it is not possible that there is a digit greater than 3 in this sequence.
[/hide]


My  contents  is  at  least  2   pages.
I  want   that   there   is   no    page   number   in   the   first   page   of    the    contents.
How to get that ?

\documentclass{report}

..........

\begin{document}

\pagenumbering{roman}
\tableofcontents

\pagenumbering{arabic}

\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg

\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg

\end{document}

Thank stevem very much!

But  if  the   contents is at least [color=red][b]2[/b][/color] pages,   \thispagestyle{empty}  doesn't work!

Actually,  It seems that  the following code works well:   there is no page number in the first page of the contents.

\documentclass{[color=red]article[/color]}
......
[color=red]\thispagestyle{empty}
\tableofcontents [/color]

But   ,      if  I   replace \documentclass{[color=red]article[/color]}   by  \documentclass{[color=red]report[/color]}  or  \documentclass{[color=red]book[/color]} ,    \thispagestyle{empty}  is  totally out of use, before or after \tableofcontents

[size=150][color=red]Suppose  that  the   contents is at least 2 pages[/color][/size]

You can't have chapters in the article class. \thispagestyle{empty} after \tableofcontents  does work.
The following document works with 3 pages in report and 5 pages in the book class and no number on first page and numbers on subsequent pages.
[code]\documentclass{report} % or \documentclass{book}
\begin{document}
\pagenumbering{roman}
\tableofcontents
\thispagestyle{empty}
\pagenumbering{arabic}
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\end{document} [/code]
Decide which class you want and stick to it.

No!

no number on  the  [color=red]last[/color] page

If  the  contents  is at least 2 pages,

try the following codes:
no number on the last page  of the contents

\documentclass{[color=red]report[/color]}
[color=red]\usepackage{hyperref} [/color]
\begin{document}

\pagenumbering{roman}
[color=red]\tableofcontents
\thispagestyle{empty} [/color]

[color=red]
\clearpage
\pagenumbering{arabic}[/color]
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg
\chapter{dd}
\section{ggg}
gggg
\section{ggg}
gg

\end{document}

\thispagestyle{empty} only applies to the current page. If the table of contents goes onto more than one page use \pagestyle{empty} instead.

If the table of contents goes onto more than one page and  I replaced \thispagestyle{empty} by  \pagestyle{empty}.
the output is no differences

[size=150][color=red]anyway

Thank stevem very, very much![/color][/size]

[code]\begin{document}
\renewcommand{\thepage}{}
\tableofcontents
\pagestyle{empty}
%\clearpage not required[/code]works for me.

\begin{document}
\renewcommand{\thepage}{}
\tableofcontents
\pagestyle{empty}
%\clearpage not required

Then ,  no number on first page and   no  numbers  on   subsequent pages.

Am  I  right?

There are no numbers on the 3 contents pages but all the following pages are numbered which is what you asked for.

I  want  that     no   number   on  the   first   page  of    the   contents   ,  but     there   are     numbers      on     subsequent       pages     of    the   contents

sorry,   my  english  is  very   bad,
I  must  work hard to study  it

If you want the first page of the contents to have no number but the second and third to be numbered ii, iii and then the subsequent pages 1,2,3,... use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=tocloft]tocloft package[/url]:
[code]\documentclass{report}
\usepackage[plainpages=false]{hyperref}
\usepackage{tocloft}
\begin{document}
\renewcommand{\cftaftertoctitle}{\thispagestyle{empty}}
\pagenumbering{roman}
\tableofcontents
\chapter{dd1}
\pagenumbering{arabic}
\section{ggg}
...[/code]

[size=200][color=red]Thank stevem very, very ,very, very, very, very ,very, very ,very, very ,very, very  much![/color][/size]

If   I replaced \documentclass{report}   by \documentclass{book},

replaced \pagenumbering{roman}   by \frontmatter,     and

replaced \pagenumbering{arabic}   by \mainmatter,

How  to  change  the  code   to   get   the    same   output,   that is ,  the first page of the contents to have no number but the second and third to be numbered ii, iii and then the subsequent pages 1,2,3,...   ?         and   The final   code will    not  get an warning(not mistake)

thank you!

How about  renewcommand  \tableofcontents?

\documentclass{book}
\usepackage{tocloft}
\tocloftpagestyle{empty}

\begin{document}

\frontmatter
\tableofcontents

\mainmatter
\chapter{foo}
...

\end{document}


Can anyone show me how an '[/b]algorithm[b]'[/b] works in creating 'computer[u] chess[/u] program[b]?'(Ex.the
solaris
[/b] type[b] that beat kasparov[/b].)I'm interested on how a
mathematician create chess program.

                                                   Respectfully,
                                                  A.Eddington

I believe that these computers just do a game-tree search with a ton of processing power to get large amounts of insight, then look at the possibilities and use human tweaked priorities to decide which one is most advantageous to follow (ex the priority of having a pawn in the middle vs having a knight out)

You may want to start off using the minimax algorithm (just to get something that works). Once you've passed that, you most definitely want to use the Alpha-Beta algorithm, and you can add lots of optimisations, using null-move forward pruning, transposition tables, history heuristics, killer moves, opening books etc.

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

Also try searching google for "chess programming", you might find some interesting info :)


I've heard that IMO is a team competition and there is poll of who's going to win.  However, I 've also heard that it is only an individual competition and the individuals get medals.  So, the team scores are unofficial, meaning that they are just people adding the scores of the contestants, nothing more.  Then, they are ranked.  So, if U.S. "wins" they don't officially win.  They just have a higher sum of scores from the people that added the scores of the individuals. Is this true? :shock:  :shock:  :shock:

Yes. That is true. It's similar to the way that the Olympic Games are for individual medals, but some people unofficially add them up and compare medal counts. The country with the most medals gets nothing.

From the sites I've been to, it is true that the teams are unofficial, but then, it's always fun to see which country had the best combined score.

yes the team results are definately unofficial but everyone still wants there team to have the most points
i know this because my dad participated in the math olympiad

[quote="interesting_move"]Hm...that's interesting.  I always thought that there was an official "team" score.  That's how they placed the teams right?

-interesting_move[/quote]

The point is that it's unofficial. They (whoever is in charge of the IMO, I guess) don't place the teams. Everyone else does.

who is in charge of IMO??? if they were from 1 of the competing countries that would be unfair and if they are not from a competing country, chances are that country isn't too mathematically inclined.

All countries send some proposals, then the guest country (and some others) choose some of these problems. The result is called ISL (Int. shortlist).
The "leaders" of all countries choose the final problems then.

leaders=coaches.

THen the assistant coaches stay with the kids till after so the coaches dont give away problems

On the official sites, there were always called leaders and deputy leaders  :wink:
But those are not exactly those that choose the problems, but approximately, they are.


function probability counting distinguishability blogs

I want to know in how many different ways can $n$ identical balls be placed into $r$ identical boxes. Consider both the cases
(i)when empty boxes are allowed
(ii)empty boxes are not allowed.

I was unable to solve this. I didn’t find it in any of the books I have so I thought to ask here. Hope that any one can help.

Please do it soon because I'm preparing for my exams.

This has come up many times:

Consider this diagram -

OO..OOXOO...OOXOO......OOXOO..OO

where there are $r-1$ X's and $n$ O's. Everything to the right of the first X represents the balls in the first box. Between the first and second X's are the balls in the second box. Etc. After the last ($r-1$th) box are the balls in the $r$th box. Hence of these $n+r-1$ symbols, we choose $r-1$ to be X's.

Thus the answer is $\binom{n+r-1}{r-1}$.

---

If empty boxes are not allowed, simply take one ball away from $n$ per box and put one in each box. Then distribute the rest, giving $\displaystyle \binom{n-1}{r-1}$.

Go Paladin!  I remember you helped me with essentially the same problem some time ago...  :)

[quote="paladin8"][quote="varun"]Paladin, what you did is I think for the case when the boxes are not identical.
Here consider the case(3 boxes, 9 balls)
Box1 gets 2, Box2 gets 3, Box3 gets 4.
And then in other arrangement
Box1 gets 4, Box2 gets 3, Box3 gets 2.

These will be counted as 2 arrangements. But, since the boxes are not identical hence we should count it only once. There will be many such pairs in the way you counted. So is that right?
Please help.[/quote]

Ah, yes I suppose this is true. So you're assuming that the boxes are not in any particular order or whatnot... hmm, that makes things sorta complicated and weird... I'll see if I can make something out of it, sorta tired right now...[/quote]
Doesn't that just make this a classic partition problem, where you want to partition the integer $n$ into a sum of $p$ smaller integers, where order of the addends is not important?  This is a pretty well-studied topic of mathematics; check it out here: http://mathworld.wolfram.com/PartitionFunctionP.html Well, actually, I guess the partition function counts divisions into all possible values of $p$, so that's not quite the same, but it is related.

[quote="probability1.01"]I'm pretty sure there is a closed form, but it is extremely messy and difficult to find.[/quote]

I agree with you, probability 1.01.

Even the problem (4) of [b]#4 [/b]which I posted is will be not easy.

kunny

[quote="probability1.01"]I'm pretty sure there is a closed form, but it is extremely messy and difficult to find.[/quote]

OK good. I'm not the only finding this problem messy then; it seems even for a single value of $n$ there is a lot of casework to do.

So, finally what is the result of this. And now I understand why in any book of senior secondary no such formula is given

Thanks guys.

ummm...
isn't this just the Stirling numbers of second kind? ;)

$S(r,n)$ ?

Hey...
Please help me with a similar problem i have been stuck for some time ....

By your problem we know the number of possible solutions to this
x_1 + x_2 + x_3 + ...+ X_r = n
0<= x_i <= n

Now tell me ->How to find a formula for number of solutions to following
x_1 + x_2 + x_3 + ...+ X_r = n
0<= x_i <= k  when k < n;

Amirhtlusa -- no.  Stirling numbers have distinguishable balls.

Imran -- your question is not analogous to the question in the original post, since your "boxes" (the variables $x_i$) are distinguishable, in which case a simple closed-form solution exists for the first question you post.  I don't know about the second one, but it seems more complicated.

JBL - Yes you are correct .. The base problem at the start of the blog maps to classic partition problem .

The problems I mentioned are much simpler .

Still is there a way to find a general formula for problem(2nd) I mentioned above .

[quote="Imran"]
Now tell me ->How to find a formula for number of solutions to following
x_1 + x_2 + x_3 + ...+ X_r = n
0<= x_i <= k  when k < n;[/quote]
I don't know whether there is a closed form, but you might give a shot at it using inclusion-exclusion principle


Prove that $ x^2\plus{}2x\plus{}q\equal{}0$ has distinct real roots if and only if $ (1\plus{}q)(x^2\plus{}2ax\plus{}q)\minus{}2(q\minus{}1)(x^2\plus{}1)\equal{}0$ does not admit real roots.

[quote="ocha"]$ (1 \plus{} q)(x^2 \plus{} 2ax \plus{} q) \minus{} 2(q \minus{} 1)(x^2 \plus{} 1) \equal{} 0$ has no real solutions for all real values $ a$?[/quote]

All parameters are reals.

I can't find out how to prove it.

Using the discriminant; $ x^2 \plus{} 2x \plus{} q$ has real solutions for $ 4\minus{}4q \ge 0 \longrightarrow q \le 1$

Expanding and simplifying the other equation yields;
$ (3\minus{}q)x^2\plus{} 2ax(1\plus{}q) \plus{} (q^2\minus{}q\plus{}2) \equal{} 0$

And this has real solutions for;
$ \left(2ax(1\plus{}q)\right)^2 \minus{} 4(3\minus{}q)(q^2\minus{}q\plus{}2) \ge 0$

This equation always holds if we set $ a$ to a very large value
Hence both equations can have real solutions for $ q \le 1, a >>0$...

where did i go wrong??


Let $ Y'$ be the set of all limit points of a set $ Y$ . Prove that $ Y'$ is closed.

I would rather have ideas on where to start, instead of a full solution. Thank you.

To prove that the complement of $ Y'$ is open, state exactly what it means that $ a\notin Y'$.

if $ a\notin Y'$ then $ a$ is not a limit point of $ Y$ so there exist an $ r>0$ such that $ d(a,b)<r$ only when $ a\equal{}b$. but i am get stuck since i do not see how that helps.

[hide="not sure if this is right"]
Let $ A\equal{}Y\cup Y'$ and it is easy to show that $ A$ is closed. So $ A^{c}$ is open and $ A^{c}\equal{}Y^{c}\cap (Y')^{c}$. Thus if $ b\in A^{c}$ then $ b$ is not in $ Y$ and is not a limit point of $ Y$. So $ b\in A^{c}\Longrightarrow b\in Y^{c}\cap (Y')^{c}\Longrightarrow b\in (Y')^{c}$.
Now since $ A^{c}$ is open there is a neighborhood $ N_{r}(b)\subset A^{c}$ so $ N_{r}(b)\subset Y^{c}\cap (Y')^{c}\Longrightarrow N_{r}(b)\subset (Y')^{c}$.
Thus $ (Y')^{c}$ is open, so $ Y'$ is closed.
[/hide]
I am not sure that what I have in the hidden text is acceptable but even if it is I would still like to see if I can generate a proof from mlok's idea.

so if $ b\notin Y'$ then $ b$ is not a limit point of $ Y$ thus there exist a neighborhood such that $ N_{r}(b)\cap Y\equal{}\phi$. so $ N_{r}(b)\subset Y^{c}$ Thus $ b$ is a limit point of $ Y^{c}$, so $ (Y')^{c}$ is the set of all limit points of $ Y^{c}$.

At this point i dont know how i would prove that $ (Y')^{c}$ is open.

Every point $ y\in N_{r}(b)$ is also not a limit point, so $ \left( Y^{\prime}\right)^{c}$ is open.

how do you get that every point $ y\in N_{r}(b)$ is also a limit point?

If $ y\in N_{r}(b)$, then $ N_{t}(y)\subset N_{r}(b)$ as long as $ t\plus{}d(y,b)\le r$. Triangle inequality.

ooh thanks. so then since every point in that neighborhood is a limit point they are all in $ Y'$ and have neighborhoods that are subsets of $ Y'$, thus making it open.
Thanks a lot.   :)

It's about the complement of $ Y'$... "every point in that neighborhood is [b]NOT[/b] a limit point"

o i mean that they are limit points of $ Y^{c}$ sorry about not making that clearer sorry.


function
inequalities
abstract algebra
real analysis
real analysis unsolved

$f(x)$ is differentiable in $(a,b)$ and continuos in $[a,b]$ and is not a straight line. Prove that there exists a $c$ in $(a,b)$ such that:
$\left| f^{'}(c) \right|>\left|\frac{f(b)-f(a)}{b-a}\right|$

[quote="perfect_radio"]Let $g : [a,b] \to \mathbb R$ such that $f(x) = g(x)+\lambda x, \, \forall x \in [a,b]$, where $\lambda = \frac{f(b)-f(a)}{b-a}$. Hence, $g(a) = g(b)$.

The conclusion is equivalent to proving there is $c \in [a,b]$ such that $\left| g^\prime (c)+\lambda \right| > \left| \frac{\left( g(b)-g(a) \right)+\lambda \left( b-a \right)}{b-a}\right| = \left| \lambda \right|$.

We suppose the opposite, i.e. $\forall x \in [a,b]$ we have that $\left| g^\prime (x)+\lambda \right| \leq \left| \lambda \right|$.

Analyzing a few cases, we discover that $g^\prime (x)$ must be either nonnegative or nonpositive.
Therefore, $g$ is an increasing (or decreasing) function such that $g(a) = g(b)$, where $a<b$. (Note that we have that only on $(a,b)$. To prove it on $[a,b]$, suppose the opposite and use Lagrange's Theorem to get a contradiction.) So $g$ is the constant function. That means that $f$ is linear, contradiction.[/quote]
Nice solution! :D I didn't think about $f$ isn't a straight line!

Nice soluTion perfect radio :) , another solution:
let $d\in(a,b)$, then if we prove that at least one of the numbers $\left|\frac{f(b)-f(d)}{b-d}\right|$ and $\left|\frac{f(d)-f(a)}{d-a}\right|$ (which none are equal to $\left|\frac{f(b)-f(a)}{b-a}\right|$)are bigger than $\left|\frac{f(b)-f(a)}{b-a}\right|$ then the problem is solved.
Suppose the opposite, so we have both inequalities:
$\left|f(b)-f(d)\right|<\frac{b-d}{b-a}\left|f(b)-f(a)\right|$
$\left|f(d)-f(a)\right|<\frac{d-a}{b-a}\left|f(b)-f(a)\right|$
 and by the sum:
$\left|f(b)-f(d)\right|+\left|f(d)-f(a)\right|<\left|f(b)-f(a)\right|$

which is contradiction.


function
integration
absolute value
calculus
calculus computations

problem: If $ f$ is constant on the interval [a,b], this it is integrable with respect to any function $ g$ and
$ \int_{a}^{b} fdg \equal{} f(a){g(b)\minus{}g(a)}$.

So it's pretty obvious by simply integrating. However, I wanted to show using Riemann -Stieltjes

$ \forall \epsilon >0, \exists \xi \in [x_{k\minus{}1},x_{k}],$ then $ |\sum_{k\equal{}1}^{n} f(\xi)[g(x_{k})\minus{}g(x_{k\minus{}1})  \minus{} f(a){g(b)\minus{}g(a)}| < \epsilon$.

since $ f$ is constant on $ [a,b]$ then

$ f(\xi)|(\sum_{k\equal{}1}^{n} [g(x_{k})\minus{}g(x_{k\minus{}1}]) \minus{} [{g(b)\minus{}g(a)}]| < \epsilon$.

I know it's obvious, but how do i show rigorously that the stuff inside the absolute value is $ < \epsilon$?

for example, could i simply say that all the sums of the partitions is exactly $ {g(b)\minus{}g(a)}$ ...and then end it at that,or do I have to be more rigorous?

First of all, it's $ \int_{a}^{b} fdg \equal{} f(a)\{g(b) \minus{} g(a)\}$. Let $ c$ be the value to which $ f$ is equal  on the interval $ [a,b]$. A R-S sum is $ c\sum_{k \equal{} 1}^{n}(g(x_k) \minus{} g(x_{k \minus{} 1})) \equal{} c(g(x_n) \minus{} g(x_0))$ (telescoping sum). No epsilons needed.


1. Determine all positive integers $n$, such that the equation $2\sin nx = \tan x + \cot x$ has solutions in real numbers $x$.

4. A line tangent to the incircle of the equilateral triangle $ABC$ intersect the sides $AB$ and $AC$ at points $D$ and $E$ respectively. Prove that     $\frac{AD}{DB} + \frac{AE}{EC} = 1$.

(1): $\tan{x}+\cot{x} \geqq 2 \geqq 2\sin{nx}$ or $\tan{x}+\cot{x} \leqq -2 \leqq 2\sin{nx}$.

[quote="yetti"][b](2)[/b] Denote a = AB = BC = CA the side length of the equilateral triangle $\triangle ABC$. We have a = AD + BD = AE + CE. The quadrilateral BCED is tangential, hence, DE + BC = BD + CE. Using the cosine theorem for the triangle $\triangle ADE$ with the angle $\angle A = 60^\circ$,

$DE^2 = AD^2 + AE^2 - AD \cdot AE$

Substituting to the equal sums of the opposite sides of the tangential quadrilateral BCED,

$\sqrt{AD^2 + AE^2 - AD \cdot AE} + a = BD + CE = 2a - (AD + AE)$

Squaring the equation,

$AD^2 + AE^2 - AD \cdot AE = a^2 + AD^2 + AE^2 - 2a(AD + AE) + 2 AD \cdot AE$

$a^2 - 2a(AD + AE) + 3AD \cdot AE = 0$

Substituting AD = a - BD, AE - a - CE,

$a^2 - 2a[2a - (BD + CE)] + 3(a - BD)(a - CE) = 0$

$a^2 - 4a^2 + 2a(BD + CE) + 3a^2 - 3a(BD + CE) + 3 BD \cdot CE = 0$

$a(BD + CE) = 3 BD \cdot CE$

$\frac{a}{CE} + \frac{a}{BD} = 3$

Substituting a = AE + CE, a = AD + BD to the respective fractions,

$\frac{AE + CE}{CE} + \frac{AD + BD}{BD} = 3$

$\frac{AE}{CE} + \frac{AD}{BD} = 1$[/quote]

The problem is to simple to think about so long solution  ;)

My solution: Let the incircle touch ED at T, AB at K and AC at L. Denote DK=x, LE=y. Let the length of the sides of triangle ABC be 1 (the situation is homogenic). Than DE=(x+y) and AD=1/2-x and AE=1/2-y. By the cosinus rule in triangle ADE we have:
$(x+y)^2=(1/2-x)^2+(1/2-y)^2-(1/2-x)(1/2-y)$ which ie equivalent to $12xy=1-2(x+y)$
On the other hand we want to prove that:
$\frac{AE}{CE} + \frac{AD}{BD} = 1$
what we can write as:
$\frac{1-2x}{2x+1} + \frac{1-2y}{2y+1} = 1$ which is equivalent  with $12xy=1-2(x+y)$, what we already know. This completed the proof.

[quote="9216"]
The problem is to simple to think about so long solution  ;)
[/quote]
This is not a proper objection, but a matter of opinion. In this case, I happen to think that the 1st of the 2 simple solutions is simpler. This is not strange, because the very purpose of opinions is to vary (see Saturnin). But I am not forcing it on you.

Yetti

[b]A easy generalization:[/b]

$\triangle ABC,\ M\in (AB),\ N\in (AC)$ and the line $MN$ is tangent to the incircle of the triangle $ABC$

$\Longrightarrow\underline {\overline {\left|\ \frac{MA}{MB}\cdot \cot \frac B2 +\frac{NA}{NC}\cdot \cot \frac C2=\cot \frac A2\ \right| }}.$


Find all values of $ x$ such that $ \displaystyle\frac{1}{x\minus{}1} \plus{} \frac{2x}{x \minus{} 1} \equal{} 5$.

hello, for $ x\ne1$ we can multiply by $ x\minus{}1$  and we get
$ 1\plus{}2x\equal{}5x\minus{}5$
$ 6\equal{}3x$
$ x\equal{}2$
Sonnhard.


integration
trigonometry
logarithms
calculus
calculus computations

[img]http://img136.imageshack.us/img136/3351/clipboard02fm6.jpg[/img]

I have no words to say  :roll:

[quote="bookworm_vn"]I have no words to say.[/quote]
Why ?! Please, one or two words there !

There is no need of Maple. A few substitutions are enough.

[quote="Carcul"][color=darkred]Find[/color] $ L\equiv\int \sqrt [3]{\tan x}\,dx$.[/quote]
[color=darkblue]I"ll ascertain at first $ I_k \equal{} \int\frac {x^k}{1 \plus{} x^k}\ \mathrm {dx}\ ,\ k\in \overline {0,5}\ .$

$ \blacktriangleright$ $ I_2 \equal{} \int\frac {x^2}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \frac 13\cdot\int\frac {\left(x^3\right)'}{1 \plus{} x^6}\ \mathrm {dx}\implies$ $ \boxed {I_2 \equal{} \frac 13\cdot\arctan\left(x^3\right) \plus{} \mathcal C}\ .$

$ \blacktriangleright I_5 \equal{} \int\frac {x^5}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \frac 16\cdot\int\frac {\left(1 \plus{} x^6\right)'}{1 \plus{} x^6}\ \mathrm {dx}$ $ \implies$ $ \boxed {I_5 \equal{} \frac 16\cdot\ln\left(1 \plus{} x^6\right) \plus{} \mathcal C}\ .$

$ \blacktriangleright I_4 \minus{} I_0 \equal{} \int\frac {x^4 \minus{} 1}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \int\frac {x^2 \minus{} 1}{1 \minus{} x^2 \plus{} x^4}\ \mathrm {dx} \equal{}$ $ \int\frac {1 \minus{} \frac {1}{x^2}}{\frac {1}{x^2} \minus{} 1 \plus{} x^2}\ \mathrm {dx} \equal{}$

$ \int\frac {\left(x \plus{} \frac 1x\right)'}{\left(x \plus{} \frac 1x\right)^2 \minus{} 3}\ \mathrm {dx}$ $ \implies$ $ \boxed {I_4 \minus{} I_0 \equal{} \frac {\sqrt 3}{6}\cdot\ln\frac {x^2 \minus{} x\sqrt 3 \plus{} 1}{x^2 \plus{} x\sqrt 3 \plus{} 1} \plus{} \mathcal C}\ .$

$ \blacktriangleright I_4 \plus{} 2I_2 \plus{} I_0 \equal{} \int\frac {\left(x^2 \plus{} 1\right)^2}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \int\frac {1 \plus{} x^2}{1 \minus{} x^2 \plus{} x^4}\ \mathrm {dx} \equal{}$ $ \int\frac {1 \plus{} \frac {1}{x^2}}{\frac {1}{x^2} \minus{} 1 \plus{} x^2}\ \mathrm {dx} \equal{}$

$ \int\frac {\left(x \minus{} \frac 1x\right)'}{\left(x \minus{} \frac 1x\right)^2 \plus{} 1}\ \mathrm {dx}$ $ \implies$ $ \boxed {I_4 \plus{} 2I_2 \plus{} I_0 \equal{} \arctan\frac {x^2 \minus{} 1}{x} \plus{} \mathcal C}\ .$

$ \blacktriangleright I_4 \minus{} I_2 \plus{} I_0 \equal{} \int\frac {x^4 \minus{} x^2 \plus{} 1}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \int\frac {1}{1 \plus{} x^2}\ \mathrm {dx}$ $ \implies$

$ \boxed {I_4 \minus{} I_2 \plus{} I_0 \equal{} \arctan x \plus{} \mathcal C}\ .$

$ \blacktriangleright I_5 \minus{} I_3 \plus{} I_1 \equal{} \int\frac {x\left(x^4 \minus{} x^2 \plus{} 1\right)}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \int\frac {x}{1 \plus{} x^2}\ \mathrm {dx}$ $ \implies$

$ \boxed {I_5 \minus{} I_3 \plus{} I_1 \equal{} \frac 12\cdot\ln \left(1 \plus{} x^2\right) \plus{} \mathcal C}\ .$

$ \blacktriangleright I_3 \plus{} I_1 \equal{} \int\frac {x(x^2 \plus{} 1)}{1 \plus{} x^6}\ \mathrm {dx} \equal{}$ $ \int\frac {x}{x^4 \minus{} x^2 \plus{} 1}\ \mathrm {dx} \equal{}$

$ \frac 12\cdot\int\frac {\left(x^2\right)'}{\left(x^2\right)^2 \minus{} \left(x^2\right) \plus{} 1}\ \mathrm {dx} \equal{}$ $ \frac 12\cdot F\left(x^2\right) \plus{} \mathcal C$, where

$ F(t) \equal{} \int\frac {1}{t^2 \minus{} t \plus{} 1}\ \mathrm {dt} \equal{}$ $ \int\frac {\left(t \minus{} \frac 12\right)'}{\left(t \minus{} \frac 12\right)^2 \plus{} \frac 34}\ \mathrm {dt}$ $ \implies$

$ F(t) \equal{} \frac {2}{\sqrt 3}\cdot\arctan\frac {2t \minus{} 1}{\sqrt 3} \plus{} \mathcal C$ $ \implies$ $ \boxed {I_3 \plus{} I_1 \equal{} \frac {\sqrt 3}{3}\cdot\arctan\frac {2x^2 \minus{} 1}{\sqrt 3} \plus{} \mathcal C}\ .$

Now we'll come back to the proposed problem. Thus,

$ \left\|\begin{array}{c} x\ : \ \equal{} \ \arctan\left(t^3\right) \\
\ \mathrm {dx}\ : \ \equal{} \ \frac {3t^2}{1 \plus{} t^6}\cdot\mathrm {dt}\end{array}\right\|$ $ \implies$ $ L\equiv\int \sqrt [3]{\tan x}\ \mathrm {dx} \equal{}$ $ F\left(\sqrt [3]x\right) \plus{} \mathcal C$,

where $ F \equal{}\int\frac {3t^3}{1 \plus{} t^6}\ \mathrm {dt}$ $ \implies$ $ \boxed {L \equal{} 3\cdot I_3\left(\sqrt [3]x\right) \plus{} \mathcal C}\ .$[/color]

That's right Virgil Nicula. Here's my method:

[hide] Start by making $ \tan x \equal{} t^3 \equal{}> x' \equal{} \frac{3t^2}{1 \plus{} t^6}$, and so

$ P \equal{} \int \sqrt[3]{\tan x}\,dx \equal{} \int t \cdot \frac{3t^2}{1 \plus{} t^6}\,dt \equal{} 3 \int \frac{t^3}{1 \plus{} t^6}\,dt$.

Now make $ t \equal{} \frac{1}{u} \equal{}> t' \equal{} \minus{} \frac{1}{u^2}$, and  $ P \equal{} \minus{}3 \int \frac{(1/u)^3}{1 \plus{} (1/u)^6} \cdot \frac{1}{u^2}\,du \equal{} \minus{}3 \int \frac{u}{1\plus{}u^6}\,dt$.

Now let's make $ u^2 \equal{} z \equal{}> u' \equal{} \frac{1}{2 \sqrt{z}}$, and so we have

$ P \equal{} \minus{}3 \int \frac{\sqrt{z}}{2 \sqrt{z} (1\plus{}z^3)}\,dz \equal{} \minus{} \frac{3}{2} \int \frac{1}{1\plus{}z^3}\,dz \equal{} \minus{} \frac{1}{2} \int \left(\frac{1}{1\plus{}z} \minus{} \frac{z\minus{}2}{z^2\minus{}z\plus{}1}\right)\,dz$

$ \equal{} \frac{1}{4} \ln \left(\frac{z^2\minus{}z\plus{}1}{(z\plus{}1)^2}\right) \minus{} \frac{\sqrt{3}}{2} \arctan \left(\frac{2z\minus{}1}{\sqrt{3}}\right) \plus{} C$

$ \equal{} \frac{1}{4} \ln \left(\frac{u^4 \minus{} u^2 \plus{} 1}{(u^2 \plus{} 1)^2}\right) \minus{} \frac{\sqrt{3}}{2} \arctan \left(\frac{2u^2 \minus{} 1}{\sqrt{3}}\right) \plus{} C$

$ \equal{} \frac{1}{4} \ln \left(\frac{t^4 \minus{} t^2 \plus{} 1}{(t^2 \plus{} 1)^2}\right) \minus{} \frac{\sqrt{3}}{2} \arctan \left(\frac{2\minus{}t^2}{\sqrt{3} t^2}\right) \plus{} C$

$ \equal{} \frac{1}{4} \ln \left[\frac{(\tan x)^{4/3} \minus{} (\tan x)^{2/3} \plus{} 1}{[(\tan x)^{2/3} \plus{} 1]^2}\right] \minus{} \frac{\sqrt{3}}{2} \arctan \left(\frac{2\minus{}(\tan x)^{2/3}}{\sqrt{3} (\tan x)^{2/3}}\right) \plus{} C$.[/hide]


linear algebra
matrix
algebra
polynomial
complex numbers
linear algebra unsolved

Nice proof Moubinool, although I'm not referee at AMM :oops:

Dan Velleman from AMM said ' it was not sure that $det(AB)=det(A)det(B)$ for $A,B$ quaternions matrices. See $(3)$ in my proof above.

That's why the paper was rejected.

Someone know some paper about determinant and quaternion matrix ?
thank's

Moubi, how do you define the determinant of a quarternion matrix?

yes i see, I need to find a definition for determinant of matrix
$M(n,H)$ where $H$ is the quaternion field (not commutative)

Hi Moubi,
  you can conclude if you show the following result:
  Let A be a (n,n) block complex matrix such that each element is a quaternion in [j,k] ie a (2,2) complex matrix:[0,a1+ia2,
-a1+ia2,0] (row by row).
   Then the characteristic polynomial of A is even and has only real coefficients.

Moubi, I think if you use the usual definition $det A = (-1)^{sgn(\sigma)}a_{1\sigma(1)}...a_{n\sigma(n)}$, then it can't be true. Example:
$A = \begin{tabular}{cc}i & j \\ i & j \end{tabular}$
$det A = 2k$, $det A^{2}= 0$

The definition of a quarternion matrix I heard is that we should first map the matrix back to complex matrix, and then compute the determinant of the complex matrix. Then $det I = 1$, $det(AB) = det(A)det(B)$ and $det(A) \neq 0$ iff matrix is invertible all holds.

Reference: Kristopher Tapps, Matrix groups for undergraduates, p.31

Um.. but with this definition the determinant of a quaternion matrix is always a complex number. As a consequence, Cramer's rule for solving linear systems fails.

There are some papers on determinants over noncommutative rings. Here is [url=http://arxiv.org/abs/math/0208146]a recent one[/url]. Not a recreational reading, but very interesting.


How many 16-bit strings are there that contain exactly 7 "1"s but with no 1's adjacent to each other?

The "first part" is easy; there are (16,7) = 11440

16-bit strings containing exactly 7 1's, but I can't figure out how to winnow out the ones that have adjacent 1's.

This is not an "open question."  The answer is $\binom{16-7+1}{7}$; it may be helpful to imagine placing the seven 1s first, then placing a 0 between any two adjacent 1s, then considering possible arrangements of the remaining 0s and seeing if this is a combinatorial problem you are familiar with.

(Actually, I like to do this slightly differently: there is a bijection between sequences of this sort and sequences of length seventeen with seven 1s, ten 0s, which end in 0.  And there is a bijection between sequences of that sort and arrangements consisting of seven 10s and three 0s.)

OK, I followed your suggestion but came up with a very different result. We start with the partial string:

1 0 1 0 1 0 1 0 1 0 1 0 1

The string contains nine 0's, with three left to place after we use six of them to "separate" the 1's. From there, I reasoned that there are 8 ways to place the first (remaining) 0 (at either end, or between any pair of 1's); then for each of these strings, there are again 8 distinct strings formed by placing the second 0; then for each of these 8^2 strings, there are 8 ways to place the final 0. This gives 8^3 possible strings rather than (10,7) = 120. Is there a problem with my reasoning?

But I have rethought the problem and come up with the same answer you gave. Rather than placing the 1's first, I placed the 9 0's with a "slot" between each, and a slot at the beginning and end. Now we have seven 1's to place in 10 slots, giving (10,7) = 120 possible strings. Problem now is, I cannot find the error in my other attempt!

Consider the situation where two of the remaining three 0s end up at the far left, while one ends up at the far right.  In how many different ways have you counted this situation?

Your last method seems nicer than mine for this problem :)

I went over the problem, and I believe that :

There are 7 1's and 9 0's.

Start by arranging the 9 0's in a line with gaps between them and a space at each end.  That gives 10 positions of which we choose 7 in which to place a 1.

These 7 positions can be chosen in

    C(10,7)  =  120 ways

So there will be 120 bit strings with 7 1's of length 16 with no
adjacent 1's.

Yes, you already said that, and it was the solution to which I was refering when I said,

"Your last method seems nicer than mine for this problem."

You also asked why you got the wrong answer trying to follow my result; my previous post pointed out an error in your reasoning.

Sorry for repeating myself. I just have a tough time with discrete math.


function
limit
advanced fields
advanced fields unsolved

suppose that $f_n \in L^p([0;1])$, $1<p<\infty$, is a sequence of functions that tends weakly towards $u \in L^p$.
Show that if $\lim |f_n|_p=|f|_p$ then $f_n$ tends strongly towards $f$.

Maybe it is easier to prove it in an arbitrary uniformly convex space  :)
And what happen for $p=1$ ..


Let $ m\leq n$ be positive integers, and let $ F$ be a family of $ m$-element subsets of $ \{1,2,...,n\}$ such that $ A\cap B$ is not empty for any $ A,B\in F$. Determine the maximum possible number of elements in $ F$.

We have two cases:
1. $ m\leq \frac{n}{2}$ : See [url=http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Ko%E2%80%93Rado_theorem]Erdos ko rado theorem[/url]

2. $ m>\frac{n}{2}$ , any two sets of cardinality $ m$ intersect, hence $ n\choose m$


Let $N$ be the positive integers set and for all  $r\in R( real)$ let  $D_{(r)}$ be the set  $D_{(r)}=\{[nr] \ |\  n \in N\} $, where $[nr] $ means "the integer part of $nr$".

Find all $x\in R$ such that $D_{(x)},D_{(x^2)}$ make a partition of $N$  :  $D_{(x)}\cap D_{(x^2)} =  \emptyset \ ,\  D_{(x)}\cup D_{(x^2)} = N$

 :cool: Who invented this problem ?!  :cool:

All we need is $\frac{1}{x}+\frac{1}{x^2}=1$. I forgot the exact name for such sequences. I  remember Pierre wrote about it many times.

I think it's Beatty sequence. Poor Pierre! :) He must have posted and proven stuff about this subject like a hundred times! :) There are some problems which always come back.. :)


inequalities
integration
limit
advanced fields
advanced fields unsolved

Let $\Omega$ be a bounded open subset of $\mathbf R^{n}$ and let $\lambda_{1}$ be the first eigenvalue of $-\Delta$ in $H_{0}^{1}(\Omega)$. Prove that the problem
\[\left\{ \begin{aligned}-\Delta u & = au+b & \text{ in }\Omega,\\ u & = 0 & \text{ on }\partial\Omega, \end{aligned}\right.\]
where $a<\lambda_{1}$ and $0<b$ are constants, has a unique solution $u$. Moreover $u>0$ in $\Omega$.

unicity follows from that fact that $a$ is not an eigenvalue of the laplacian, so $-\Delta-aId$ is injective (in fact it is an isomorphism from $H^{1}_{0}(\Omega)$ to $H^{-1}(\Omega)$).

if $u_{-}= 1_{\mathbb{R}^{-}}\circ u$, one has to prove that $0=u_{-}\in H^{1}_{0}(\Omega)$. Hence (Poincaré inequality) it suffices to prove that $\nabla u_{-}= 0$. The case where $a \leq 0$ is easy because:
$0 \leq (b|u_{-}) = (-\Delta u-a u | u_{-}) = \ldots =-(\nabla (u_{-}) | \nabla(u_{-}))+a (u_{-}| u_{-})$ hence $(\nabla(u_{-}) | \nabla(u_{-})) = 0$ because $a \leq 0$. What about the general case ?

Good proof. Here's mine (I believe there're lots of others).

Consider a functional
\[I(u) = \frac{1}{2}\int_\Omega |\nabla u|^{2}-\frac{a}{2}\int_\Omega u^{2}-b\int_\Omega u, \]
then $u$ is a solution iff it is a critical point of $I$ ($I'(u)=0$). You have already proved the uniqueness of such a point.

$I$ is weakly lower semicontinuous on $H_{0}^{1}(\Omega)$ and, since $a<\lambda_{1}$ and $0<b$, it is coercive ($\lim_{|u|\to\infty}I(u) =+\infty$), hence $I$ attains its minimum. Thus $u$ is the solution iff it is the minumum of $I$. But $I(|u|) \le I(u)$ hence, by uniqueness, $u=|u|\ge0$.

Thus $-\Delta u = au+b \ge b > 0$ in $\Omega$ and, by maximum principle, $u>0$ in $\Omega$.

nice proof  :) in the case $a>0$, prove that $\|u\|_{\infty}\leq \frac{b}{a}$ ..

If $a\le0$, then $-\Delta u+|a|u=b>0$ and we can again apply (strong) maximum principle.  :)


trigonometry
geometry
perimeter
inequalities
Euler
trig identities
Law of Sines

Find the maximum value of \[ \sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} \] where $\alpha,\beta$ and $\gamma$ are positive and $\alpha + \beta + \gamma = 180^{\circ}$.

Same thing as a + b + c = pi/2 and we want sina + sinb + sinc maximized.

Now sin x is concave down in [0, pi/2], so by Jensen's we have the result $\sum \sin a \leq 3/2$ immediately.

[quote="Singular"]Same thing as a + b + c = pi/2 and we want sina + sinb + sinc maximized.[/quote]

I don't think so...

Max is $\dfrac{3}{2} \sqrt{3}$

Probably there exists tens of proofs :

A taste of trigonometric gives :
 $\sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} = 4 \ \sin{\alpha} \sin{\beta} \sin{\gamma}$

Then AM-MG inequality gives $\sin{\alpha} \sin{\beta} \sin{\gamma} \leq {(\sin{\alpha} + \sin{\beta} +\sin{\gamma} ) / 3)}^3$
Then $\sin{\alpha} + \sin{\beta} +\sin{\gamma} \leq 3/2 \sqrt{3}$ is well known (using Jensen for instance).

a taste of geometry gives (law of sines ..) :
$\sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} = 1/R (a \cos{\alpha} + \cos{\beta} +\cos{\gamma}) ) = 2A/R^2$
since
$\sin{\alpha} + \sin{\beta} + \sin{\gamma} = A/rR$

we obtain the nice equality :
$\sin{\alpha} + \sin{\beta} + \sin{\gamma} / (\sin{2\alpha} + \sin{2\beta} + \sin{2\gamma}) = R/2r \geq 1$ by Euler inequality.

[quote="shobber"][quote="Gyan"]We need to find the maximum value of the perimeter of a triangle inscribed in a unit circle.[/quote]
Can you make a proof of it?[/quote]
Find the maximum value of the sum of chords in a circle if one side of a triangle is fixed. And this is a standard olympiad question. If you don't know then take this.
Chord AB subtends an arc. Find a point P on the arc such that AP + BP is maximum.
Moreover you have a wonderful Butterfly even with you to think about such elementary things.


calculus
integration
function
derivative
logarithms
complex analysis
real analysis

Not an easy problem, I think:
  Let $f$ be a continuous real function defined on $[0,a]$ which is non-zero (that is, $f$ is not the identical 0 function). Prove that the sequence $\int_{0}^{a}{ e^{nx}f(x)dx}$ is unbounded.

No, it is not easy at all. The only solution I found uses heavy complex analysis tools (sorry, Harazi  :P ). It goes as follows. Define $F(z)=\int_0^a e^{zx}f(x)\,dx$. Clearly, $F(z)$ is well defined for all $z\in\mathbb C$ and is an entire function. Let us observe that $F$ is not identically $0$ if $f$ is not identically $0$ (one reason for this is that the restriction of $F$ to the imaginary axis is just the Fourier transform of $f$). Also $F(z)$ and all its derivatives are  uniformly bounded in the left half-plane $\mathbb C_-=\{\text{Re}\,z\leqslant 0\}$ and grows not faster than $e^{a\text{Re\,}z}$ in the right half-plane $\mathbb C_+$. At last, $F(z)\to 0$ as $\text{Re\,}z\to-\infty$. From here it follows immediately that $F$ cannot be bounded on the positive semiaxis. Indeed, if it were bounded there, we could use the Phragmen-Lindelof principle to conclude that $F$ is bounded in the right half-plane and, thereby, on the entire complex plane, which would force it to be a constant and, therefore, identically $0$.

Unfortunately, this is not exactly what we want because our aim is to show that $F$ is unbounded on the set of positive integers. But it gives one an idea of how to proceed. Assume that $F$ is bounded on the set of all positive integers. Our first task will be to show that then $F$ is bounded on the positive semiaxis outside, maybe, a small exceptional set. To this end, assume that $|F(n)|\leqslant 1$ for every positive integer $n$. Since $F$ is real on the positive semiaxis $\mathbb R_+$, every interval $[n,n+1]$ on which $F$ is not bounded by $1$ contains a zero of $F'$. Since $F'$ is bounded on the imaginary axis and grows at most exponentially in the right half-plane, the positive real zeroes $x_j$ of $F'$ satisfy the Blaschke condition $\sum_j \frac 1{x_j}<+\infty$. This implies that $F$ is bounded by $1$ on $\mathbb R_+\setminus E$ where $E$ is the union of all intervals $[n,n+1]$ containing one of $x_j$ and $\int_E \frac {dx}x<+\infty$ (we assume that $F$ is bounded by $1$ on $[0,1]$, which can be easily ensured by multiplying $F$ by a small positive constant).

The next step is to show that boundedness of $F$ on $\mathbb R_+\setminus E$ together with the a priori growth restriction $|F(z)| \leqslant e^{a\text{Re\,}z}$ in the right half-plane implies that $F$ grows slower than any exponent on $e^{i\frac \pi 4}\mathbb R_+$. There is a version of Phragmen-Lindelof principle that will allow us to conclude from there that $F$ is bounded in the right half-plane. Consider the function $H(z)=F(\sqrt z)$ in the upper half-plane. Note that $H$ is bounded by $1$ on $\mathbb R\setminus G$ where $G= E^2=\{x^2\,:\,x\in E\}$. Also $H(z)$ grows only as $e^{a\sqrt{|z|}}$ in the upper half-plane. It follows that $\log |H(iy)|\leqslant \frac 1\pi\int_{\mathbb R}\log |H(x)|\frac y{x^2+y^2}\,dx\leqslant \frac a\pi\int_{G}\sqrt x\frac y{x^2+y^2}\,dx=I(y)\sqrt y$ where $I(y)=\frac a\pi\int_{G}\sqrt {\frac xy}\frac y{x^2+y^2}\,dx$. Note now that the integrand in $I(y)$ is dominated by $\frac 1x$ (just consider the cases $x<y$ and $x>y$ separately) and tends to $0$ for fixed $x$ as $y\to+\infty$. Thus, using the dominated convergence theorem (recall that $\int_G\frac {dx}x=2\int_E\frac {dx}x<+\infty$), we conclude that $I(y)\to 0$ as $y\to+\infty$, which is equivalent to what we need.

The end. As usual, questions are welcome :) .

Oh, my god!!!  :(  :(  :( . The first part was easy to understand till the principle of the two men. After that... Interesting, the first indication of the examinator was to prove a monstruous identity of Phragmen. After that the problem was easy. I don't remember now the identity (of course!!!), but if you are interested I will post it. It gives a nice elementary solution.

I have this problem in a book by Sorin and Marius Radulescu, and the problem before this one is that ugly identity by Phragmen :). I just thought it was a nice coincidence, and it was worth mentioning it :).

Guys please post that Phragmen identity.

Here it goes: for any real continuous function $f$ defined on $[0,a]$ we have that the limit when $x$ is very large of  $\sum_{k\geq 1}{\frac{(-1)^{k-1}}{k!}\cdot\int_{0}^{a}{e^{kx(t-u)}}f(u)du}=\int_{0}^{t}{f(u)du}$ for all $t$ in $[0,a]$.


The sum of the reciprocals of the first four consecutive positive integers is greater than two.  What is the least number of consecutive positive integers necessary to make the sum of the reciprocals greater than three?

[hide]11--It's too hard to write right now.[/hide]

[quote="abacadaea"]please dont review old topics. if youre new to AoPS, then read the anouncements befroe you post anything[/quote]

OK!!  :agent: