## Search found 217 matches

- Thu Feb 28, 2013 1:42 pm
- Forum: Algebra
- Topic: Vietnam 2012- Surjective function
- Replies:
**4** - Views:
**2064**

### Re: Vietnam 2012- Surjective function

There are two cases.I missed it.One case gives $a=0$ and $b=x$ another cases gives $b=0$ and $a=x$.One case gives no solution while the other case gives $f(x)=4x$

- Thu Feb 28, 2013 3:46 am
- Forum: Algebra
- Topic: AN INTERESTING PROBLEM BY SAKAL DA
- Replies:
**5** - Views:
**2501**

### Re: AN INTERESTING PROBLEM BY SAKAL DA

Let $b_n=a_n-2012$ notice that $a_n$ depends totally on $a_{n-1}$.$a_{n-1}-2012=a_n(a_n-2012)$

So $b_{n+1}=b_n(b_n+2012)$

Then the general formula can be established using induction.I see no better way right now.

So $b_{n+1}=b_n(b_n+2012)$

Then the general formula can be established using induction.I see no better way right now.

- Thu Feb 28, 2013 3:16 am
- Forum: Algebra
- Topic: Vietnam 2012- Surjective function
- Replies:
**4** - Views:
**2064**

### Re: Vietnam 2012- Surjective function

$f(f(0))=f(0)$ and strictly increasing property yields $f(0)=0$.

let $a_n=f^{n}(x)$ and let $a_0=x$

then $a_n=ap^n+bq^n$ where $a$ and $b$ are the solutions of $x^2-x-12=0$

$a_2=f(x)=ap+bq=-3p+4q=-3x+7q$

but $f(0)=0$ implies $q=0$

so $f(x)=-3x$ which is surjective and strictly increasing.

let $a_n=f^{n}(x)$ and let $a_0=x$

then $a_n=ap^n+bq^n$ where $a$ and $b$ are the solutions of $x^2-x-12=0$

$a_2=f(x)=ap+bq=-3p+4q=-3x+7q$

but $f(0)=0$ implies $q=0$

so $f(x)=-3x$ which is surjective and strictly increasing.

- Thu Feb 28, 2013 3:02 am
- Forum: Algebra
- Topic: exponential equation.
- Replies:
**4** - Views:
**2037**

### Re: exponential equation.

The given equation splits up into $2$ different equations. $2^x=x$ for positive $x$. and $2^x=-x$ for negative $x$. That means the equation we have to solve is this $2^x=|x|$ When $x$ is negative the left side is less than $1$ and tending to $0$.So no solution in this case. If $x$ is positive the gr...

- Thu Feb 28, 2013 2:50 am
- Forum: Algebra
- Topic: find all polynomial
- Replies:
**1** - Views:
**1295**

### Re: find all polynomial

$0$ is obvoiusly a root.Which makes $1$ to be a root that means $2$ is another root......................at last $25$ is a root.

So $P(x)=x(x-1)(x-2).................(x-25)Q(x)$

Plugging this into the original equation $Q(x-1)=Q(x)=c$

So $P(x)=cx(x-1)(x-2)............(x-25)$

So $P(x)=x(x-1)(x-2).................(x-25)Q(x)$

Plugging this into the original equation $Q(x-1)=Q(x)=c$

So $P(x)=cx(x-1)(x-2)............(x-25)$

- Thu Feb 28, 2013 2:27 am
- Forum: Algebra
- Topic: N'th Differencial
- Replies:
**3** - Views:
**1743**

### Re: N'th Differencial

Induction and combinatorial argument both yeilds easy solution.

- Thu Feb 28, 2013 2:24 am
- Forum: Algebra
- Topic: Find value using the roots of polynomial
- Replies:
**5** - Views:
**3450**

### Re: Find value using the roots of polynomial

Let $S=\frac{\alpha-1}{\alpha+1}+\frac{\beta-1}{\beta+1}+\frac{\gamma-1}{\gamma+1}$

$S+3=2 (\frac{\alpha}{\alpha+1}+\frac{\beta}{\beta+1}+\frac{\gamma}{\gamma+1})$

$S+3=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=2$

I posted in the 1st topic of algebra forum.

$S+3=2 (\frac{\alpha}{\alpha+1}+\frac{\beta}{\beta+1}+\frac{\gamma}{\gamma+1})$

$S+3=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=2$

I posted in the 1st topic of algebra forum.

- Thu Feb 28, 2013 2:11 am
- Forum: Algebra
- Topic: Polynomial equation
- Replies:
**5** - Views:
**2556**

### Re: Polynomial equation

$(x^2+5x+4)(x^2+5x+6)=120x^2$

$\Longrightarrow (x^2+5x+5)^2-1=120x^2$

$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$

$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$

so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.

$\Longrightarrow (x^2+5x+5)^2-1=120x^2$

$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$

$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$

so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.

- Wed Feb 27, 2013 7:38 pm
- Forum: Algebra
- Topic: Uzbekistan TST 2012-PROBLEM-2
- Replies:
**7** - Views:
**2744**

### Re: Uzbekistan TST 2012-PROBLEM-2

Nayel vai's solution is more than impressive.

- Wed Feb 27, 2013 12:10 pm
- Forum: Algebra
- Topic: Uzbekistan TST 2012-PROBLEM-2
- Replies:
**7** - Views:
**2744**

### Re: Uzbekistan TST 2012-PROBLEM-2

$f(x)=f(a) \frac{(x-b)(x-c)}{(a-b)(a-c)}+f(b) \frac{(x-a)(x-c)}{(b-a)(b-c)}+f(c) \frac{(x-a)(x-b)}{(c-a)(c-b)}$

Now plugging $x=a+b+c$ we get $f(a+b+c)$ .

Now plugging $x=a+b+c$ we get $f(a+b+c)$ .