How do I find the Taylor polynomial about x=0 for the following:
(sin(x) cos(x))/x ?
The answer is: 1 - 4x^2 / 3! + 16x^4/5!
I tried working it out by first rewriting the function as sin2x / 2x and got the first derivative as (2xcos2x - sin2x)/(2x^2).
But then realized all subsequent derivatives after the first will have some power of x in the denominator and so won't all the terms be 0?
The only way this works is if I take 1/2x(taylor expansion of sin2x)
Finding the Taylor polynomial series for sinxcosx/x watch
- Thread Starter
- 25-03-2013 19:55
- 25-03-2013 19:59
THe original equation isn't defined at x=0 so you cannot taylor expand about it.
- 25-03-2013 20:23