Search found 153 matches
- Thu Feb 23, 2012 10:29 am
- Forum: Algebra
- Topic: Continuous function: Q <--> R\Q
- Replies: 9
- Views: 5067
Re: Continuous function: Q <--> R\Q
if co-domain of $g(x)=x+f(x)$ have two distinct element $a,b$ then it's co-domain also have a rational number between $a$ and $b$ by mean value theorem as $g$ is continuous. so, $g$ is constant and $f(x)=c-x$ where $c$ is a irrational number. But then $f(\frac{c}{2})=\frac{c}{2}$ which is a contradi...
- Mon Feb 13, 2012 8:40 pm
- Forum: Combinatorics
- Topic: Probability: coin toss
- Replies: 7
- Views: 4974
Re: Probability: coin toss
Let $C_n$ is the number of such sequence, Let, in a such sequence there are $k$ Heads $a_0-1$ is the number of Tails before the first Head $a_i$ is the number of Tails between the $(i-1)$'th and $i$'th Head, $0<i<k$ $a_k-1$ is the number of Tails after the $k$'th Head It's easy to see that, $a_i>0$ ...
- Wed Feb 01, 2012 9:31 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: A proposed problem of IMO
- Replies: 12
- Views: 8024
Re: A proposed problem of IMO
@Adib: Your Induction hypothesis is not enough to prove the problem, it just proves a case of the given problem. Note that, the problem asked you to prove for all $c_i$ with only one condition $(n-1)\sum_{i\le n}{c_i^2} = \left ( \sum_{i\le n}{c_i} \right )^2$. That is, you can't add a extra constra...
- Tue Jan 31, 2012 11:02 pm
- Forum: Number Theory
- Topic: Sequences of interest
- Replies: 5
- Views: 3551
Re: Sequences of interest
Yes! Let, $[x]$ denotes $x$ rounded to closest integer. Then it is easy to find, $[x] = \lfloor x + \frac{1}{2} \rfloor$ And, another obvious thing is, $\lceil x \rceil = \lfloor x \rfloor +1$ iff $x \not\in \mathbb{Z}$ It can also be easily found that, $\left[\sqrt{n}\right]=\left[\sqrt{n-\frac{1}{...
- Thu Jan 26, 2012 3:17 pm
- Forum: Number Theory
- Topic: Sequences of interest
- Replies: 5
- Views: 3551
Re: Sequences of interest
\[ G(n) = \left\lceil n-\frac{1}{2}+\sqrt{n-\frac{1}{2}}\right\rceil \]
- Wed Jan 25, 2012 10:29 pm
- Forum: Combinatorics
- Topic: Digit-swapping game (own)
- Replies: 5
- Views: 4424
Re: Digit-swapping game (own)
This can be generalized for all natural number $n$. Let, for any permutation $(a_1,a_2,\cdots,a_n)$ of $(0,2,\cdots, n-1)$, $f_n(a_1,a_2,\cdots,a_n) = \sum_{n > i \ge 0} {\left|\{\ j \ |\ j<i, a_j < a_i\ \}\right|}$ Then, it can be ( not so maybe ) easily verified that, $f_n(\cdots,a_{i-1},a_i,a_{i+...
- Mon Jan 09, 2012 8:08 pm
- Forum: Computer Science
- Topic: factorial
- Replies: 4
- Views: 5132
Re: factorial
You can input a large number as a string and store it in a array. It'll be helpfull if you know struct (just Google) Here's a sample elementary code ( in C++, I don't know C that much :? ) : #define MAXS something // 'something' is the maximum number of digits you need struct BigInt { int digit[MAXS...
- Sat Jan 07, 2012 1:03 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2011 problem-5
- Replies: 5
- Views: 4461
Re: APMO 2011 problem-5
I asked why $f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0$? and why $x\longrightarrow 2x\Longrightarrow f(xf(y)) \leq xf(y)$ ?
- Sat Jan 07, 2012 3:03 am
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2011 problem-5
- Replies: 5
- Views: 4461
Re: APMO 2011 problem-5
*Mahi* wrote: $f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0 $
Why ? :-/*Mahi* wrote: So, setting $x$ instead of $2x$, $f(xf(y)) \leq xf(y)$, and thus $yf(x) \geq f(xy)$.
- Thu Jan 05, 2012 2:43 pm
- Forum: Computer Science
- Topic: প্রোগ্রামিং ক্যাম্প
- Replies: 2
- Views: 2900