evaluate Tn x at a4 when fx1x1Solution 1x 1 1 x1 1

Solution

1/x = 1/ [ 1 - ( - [x-1] ) ] = 1 + [-(x-1)] + [-(x-1)]^2 + [-(x-1)]^3 +.. (where we let u = -(x-1) in the geometric series) Or 1/x = 1 - (x-1) + (x-1)^2 - (x-1)^3 + ... Now integrate both sides: ln(x) = C + x - (x-1)^2/ 2 + (x-1)^3/ 3 - (x-1)^4/ 4 + .... Where C is the constant from taking the antiderivative. Substituting x=1 we get: ln(1) = C + 1 - 0 + 0 -...; since ln(1) = 0, we see that C= -1. Associating this value of C with the first term of the series (which is x) we get: ln(x) = (x-1) - (x-1)^2/ 2 + (x-1)^3/ 3 - (x-1)^4/ 4 + .... This is the Taylor\'s series for ln(x) centered at x=1. Now the \"Taylor Polynomial\" is not exactly the same as the whole infinite series. The n th degree Taylor polynomial (written Tn(x) ) is simply the sum of the terms up to the n th power. In this case, we want the 4 th degree Taylor polynomial of ln(x) at x=1 which means the answer is : T4(x) = (x-1) - (x-1)^2/ 2 + (x-1)^3/ 3 - (x-1)^4/ 4-1