I know how to aproximate it, is it enough to actually aproximate it to prove it can be aproximated?

I just noticed there is a suggestion here at the end. Sorry for lateness

Do Taylor arround the funcion $f(t):=t\sqrt{1-t}$ and $t\in [-1,1]$ Show that $f\in C^\infty$ in $[-1,1)$ and that its taylor serioes is uniformly convergent in compacts of $[-1,1)$ Its derivate tends to $\infty$ if $t\rightarrow +1$ but f is continuous, then it is uniformly continuous in $ [-1,+1]$.

Not sure which theorem or result should be used here.

  • $\begingroup$ Well... of course. If you want to prove something can be done, then actually doing it is one way of proving that. But you have to actually show you did it. In your case, you need to prove that you actually made an approximation. $\endgroup$ – 5xum Apr 18 '16 at 7:15
  • $\begingroup$ because $\text{sign}(x)$ can be approximated at an arbitrary precision (on $[-1,1]$) by some polynomials. $\endgroup$ – reuns Apr 18 '16 at 7:15
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    $\begingroup$ @user1952009 Not uniformly on $[-1,1]$. $\endgroup$ – user228113 Apr 18 '16 at 7:16
  • $\begingroup$ @joséosorio Is your approximation uniform? $\endgroup$ – user228113 Apr 18 '16 at 7:16
  • $\begingroup$ @G.Sassatelli : yes of course, that was left as the exercice (to see that it is an approximation in $L^p$ norm) $\endgroup$ – reuns Apr 18 '16 at 7:17

With Stone-Weierstrass theorem, since $|x|$ is continuous on a closed inteval, it can be approximated uniformly by polynomials. If you want an example of such an approximation of your function, you can use Berstein's polynoms :$$B_n(f)=\sum_{k=0}^n f\left(\frac{k}{n}\right)\binom{n}{k}x^k(1-x)^{n-k}$$

  • $\begingroup$ Oh you are right i got this $\endgroup$ – José Osorio Apr 18 '16 at 7:35

You'd need to prove that the approximation you find is an approximation.


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