How do I approximate #sqrt(128)# using a Taylor polynomial centered at 125?

1 Answer
Mar 8, 2015

To approximating #f(x)=sqrtx# by a Taylor polynomial centered at #125# , you'll need to know #sqrt125#.

The Taylor polynomial, of degree #n+1#, to approximate #f(x)# centered at #a# involves:

#f(a), f'(a), f''(a), f'''(a), . . . , f^((n+1))(a)#

and also involves #x-a#

Derivatives of #sqrtx# of every order will involve #sqrt(x)^k#, Thus, every term will require #sqrt125#

Re-read the question you were asked.

The exercise would make more sense if either:
(a) the question is to approximate #root(3)128#
(#f(x)=root(3)x# and its derivatives are not too hard to find when #x=125#

or

(b) #a# is not specified
(In which case, note that #sqrt128=sqrt(2^7)=2^(7/2)# so you can use #f(x)=x^(7/2)# centered at #1# to approximate #f(2)#)