Help with a Taylor series problem!!

[cindy174]

This is problem #83 on the 1998 AP Calculus BC test:

The Taylor series for lnx, centered at x=1, is sum{n=1 to infinity}[(-1)^(n+1) * (x-1)^n / n]. Let f be the function given by the sum of the first three nonzero terms of this series. Find the maximum value of abs( lnx - f(x) ).

The answer is given to be 0.145

How do you solve this?

 

[skeeter]

get out your graphing calculator ...

Y1 = (x-1) - (x-1)^2/2 + (x-1)^3/3
Y2 = ln(x)
Y3 = abs(Y2 - Y1)

graph Y3 in the interval given in the question (which, btw, you forgot to mention in your post)

determine the max.

 

[cindy174]

Taylor Series

Sorry. Thanks for the reply.

the interval is [.3,1.7]

I was wondering if there is a possible analytic solution.

 

[skeeter]

I was wondering if there is a possible analytic solution.

yes ...

how do you find the max or min of any function on an interval?

let h(x) = |ln(x) - f(x)|

h'(x) = {[ln(x) - f(x)]/|ln(x) - f(x)|}*|(1/x) - f'(x)|

analyze h'(x) over the given closed interval to locate extrema.