So, I have the series of [itex]g(x) = e^{(x-1)^{2}} = 1 + (x-1)^{2} + \frac{(x-1)^{4}}{2} + \frac{(x-1)^{6}}{6} + ... + \frac{(x-1)^{2n}}{n!} [/itex](adsbygoogle = window.adsbygoogle || []).push({});

and I am asked to find the series of [itex]f(x) = \frac{e^{(x-1)^{2}}-1}{(x-1)^{2}}[/itex] for x [itex]\neq[/itex] 1 and f(1) = 1. The Taylor series is centered about x = 1

I am told that there is an easy way to do this, but I don't see it.

any help would be appreciated, thanks

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

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!

# Shortcut to taylor series of f, given taylor series of g

Loading...

Similar Threads for Shortcut taylor series | Date |
---|---|

I Taylor expansion of f(x+a) | Nov 1, 2017 |

Shortcut for Quotient Rule | Dec 9, 2012 |

Chinese integration shortcut? | May 3, 2012 |

Name for integration by parts shortcut | Oct 13, 2009 |

Needed shortcuts | Nov 18, 2007 |

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