approximation

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

Prove or give some ideas how to do it(please):
$min_{R(x)=\frac{P(x)}{(x^2+\alpha^2)^n}}||(|x|-R(x))||_{[-1,1]}\sim {\frac{lnn}{n^2}}$ , where min takes over all rational functions of the form
$R(x)=\frac{P(x)}{(x^2+\alpha^2)^n}$ , where $P(x)$ -polynomial of degree $2n$ , and $||f(x)||_{[-1,1]}=max_{[-1,1]}|f(x)|$ -usual norm in $C[-1,1]$

Myth

Do you mean $P(x)$ is apolynomial of degree $n$ ? And $\alpha$ is a fixed number?

**Posted:** Sun Feb 27, 2005 10:58 pm

_________________
Myth is out of here

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

I'm sorry-I've edited it. I'd like to remind that minimum takes over all rational function of the form,which was desribed above- together with $\alpha$ -that's means that $\alpha$ -is not fixed.

**Posted:** Mon Feb 28, 2005 9:47 pm

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

maybe any ideas...???

**Posted:** Sun Mar 06, 2005 2:09 am

fedja

Well, since any Lipschitz function can be approximated by polynomials at the rate $1/n$ on any compact interval, the statement of the problem is still wrong...

**Posted:** Sun Mar 06, 2005 5:57 am

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

you see that we are approximating using not polynomials but rational functions with only one conjugate pole, and in our case there could be another answer

**Posted:** Sun Mar 06, 2005 9:41 pm

fedja

Since $\alpha$ is free, we can let it go to $\infty$ making the denominator essentially constant, so the speed of approximation by your rational functions cannot be worse than by polynomials...

**Posted:** Mon Mar 07, 2005 3:59 am

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

i'm sorry; it must be $\frac{lnn}{n^2}$ -i've already edited it

**Posted:** Mon Mar 07, 2005 10:40 pm