Advanced Calculus - Differentiable and Converging Polynomials

[bradyrsmith31]

I could really use some help on this problem as well!

Thanks!

 

[Fallen Angel]

Hi,

For a better help, it's useful that you post what you have tried.

Maybe in this exercise it suffices to tell you that $F(x)=arctan(x)$

 

[chisigma]

I could really use some help on this problem as well!

Thanks!

For -1 < x < 1 the following MacLaurin expansion holds...

$\displaystyle \frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} (-1)^{n}\ x^{2 n}\ (1)$

The series (1) converges uniformly, so that You can apply the series integration theorem and write...

$\displaystyle F(x)= \int_{0}^{x} \frac{d t}{1 + t^{2}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2 n + 1}}{2 n + 1}\ (2)$

The series expansion (2) has only terms of odd degree, so that F(x) is an odd function and is $\displaystyle F(- x) = - F(x)$. The series (2) is also 'alternating signs', so that for n even is $\displaystyle P_{n} (x) > F(x)$ and for n odd is $\displaystyle P_{n} (x) < F(x)$. Finally the (2) in [0,1] converges to a continuos function, so that it converges uniformly...

Kind regards

$\chi$ $\sigma$