I ijd5000 Junior Member Joined Sep 3, 2013 Messages 51 Dec 14, 2013 #1 Find the Taylor Series centered at x=pi for f(x)=cos(x) i made the table of derivatives evaluated at a nf^(n)(x)f^(n)(pi)0cos(x)-11-sin(x)02-cos(x)13sin(x)04cos(x)-1 I'm having trouble expressing the patter for the numerator of the series
Find the Taylor Series centered at x=pi for f(x)=cos(x) i made the table of derivatives evaluated at a nf^(n)(x)f^(n)(pi)0cos(x)-11-sin(x)02-cos(x)13sin(x)04cos(x)-1 I'm having trouble expressing the patter for the numerator of the series
I ijd5000 Junior Member Joined Sep 3, 2013 Messages 51 Dec 14, 2013 #2 Romsek said: \(\displaystyle (-1)^{(k+1)}x^{2k}\text{ because you only use even powers of x.}\) \(\displaystyle \text{and likewise }(2k)!\text{ for the}\) denominator. Click to expand... why will the denominator be (2k)! ? also since it's a taylor series shouldn't it be (x-pi)^2k ?
Romsek said: \(\displaystyle (-1)^{(k+1)}x^{2k}\text{ because you only use even powers of x.}\) \(\displaystyle \text{and likewise }(2k)!\text{ for the}\) denominator. Click to expand... why will the denominator be (2k)! ? also since it's a taylor series shouldn't it be (x-pi)^2k ?
D Deleted member 4993 Guest Dec 14, 2013 #3 ijd5000 said: why will the denominator be (2k)! ? Click to expand... Because - only the even terms survive (non-zero term). So the 2nd, 4 th , 6 th, etc. terms survive - which can be expressed as 2k terms for k = 1, 2, 3..... Similarly odd terms (1, 3, 5, etc.) can be expressed as (2k + 1) terms for k = 1, 2, 3..... Understand this "short-cut" well - because you'll see these many times....
ijd5000 said: why will the denominator be (2k)! ? Click to expand... Because - only the even terms survive (non-zero term). So the 2nd, 4 th , 6 th, etc. terms survive - which can be expressed as 2k terms for k = 1, 2, 3..... Similarly odd terms (1, 3, 5, etc.) can be expressed as (2k + 1) terms for k = 1, 2, 3..... Understand this "short-cut" well - because you'll see these many times....
