Hello,(adsbygoogle = window.adsbygoogle || []).push({});

I want to prove that the taylor expansion of [itex]f(x)={\frac{1}{\sqrt{1-x}}}[/itex] converges to ƒ for -1<x<1. If I didn't make a mistake the maclaurin series should look like this:

[itex]Tf(x;0)=1+\sum_{n=1}^\infty{\frac{(2n)!}{(2^n n!)^2}}x^n[/itex]

My attempt is to use the lagrange error bound, which is given by [itex]E(x)\leq M\frac{|x|^{n+1}}{(n+1)!}[/itex], where M is the upper bound for [itex]|f^{(n+1)}(x)|[/itex] for the interval [0,x] and show it tends to zero for [itex]\lim_{n\rightarrow \infty}[/itex].

My problem is the following: For functions like e^{x}or sin(x) M is independent of n and thus the limit is easy to compute, but here M is given by [itex]f^{(n+1)}(x)=\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}[/itex] (which I conclude from graphing the function). So I need to prove that the following holds true (for -1<x<1):

[itex]\lim_{n\rightarrow \infty}(\frac{(2n)!}{4^n n!}(1-x)^{-\frac{2n+1}{2}}\frac{|x|^{n+1}}{(n+1)!})[/itex]=0

I have no clue how to solve such a limit and would appreciate any help! Of course I don't expect anyone to work it out, but maybe someone has an advice as how to tackle complicated limits like this one... .

Thanks in advance!

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proof Taylor series of (1-x)^(-1/2) converges to function

Loading...

Similar Threads - Proof Taylor series | Date |
---|---|

B Proof of a limit rule | Dec 19, 2017 |

Would this require a Taylor Series Proof | Oct 19, 2012 |

Proof taylor formula | Nov 20, 2011 |

Proof of Taylor's formula for polynomials | Apr 12, 2009 |

Proof of taylor series | Jan 22, 2007 |

**Physics Forums - The Fusion of Science and Community**