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 July 10th, 2018, 09:04 AM #1 Newbie   Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 Taylor series of multiplied functions I am having hard times trying to write the Taylor series of: x*(cos(2pi*x)) to 2nd order around the point a=1 What do I need to do? thanks July 10th, 2018, 09:43 AM   #2
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 Originally Posted by Mathmatizer I am having hard times trying to write the Taylor series of: x*(cos(2pi*x)) to 2nd order around the point a=1 What do I need to do? thanks
There are two ways to do this. First, simply find the Taylor series for $\displaystyle x~\cos( 2 \pi x) \approx 1 + (x - 1) + 2 \pi (x - 1)^2 + \text{ ... }$

The other way is to find the Taylor series of each individual function:
$\displaystyle x = 1 + (x - 1)$

$\displaystyle \cos( 2 \pi x) \approx 1 + 2 \pi^2 (x - 1)^2 + \text{ ... }$

Multiply the two expressions and discard the term in (x - 1)^3 and you'll get the expression for the whole thing given above.

-Dan

Last edited by skipjack; July 10th, 2018 at 11:12 AM. July 10th, 2018, 09:46 AM #3 Newbie   Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 Thanks, I was able to solve it! Tags functions, multiplied, series, taylor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post chuackl Calculus 1 June 1st, 2014 06:47 PM g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM goldenarms Calculus 1 November 12th, 2009 01:36 PM vjj Number Theory 0 December 31st, 1969 04:00 PM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

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