Let Tnx be the Taylor Polynomial for fxlnx at a1 and c1 Show

Solution

In a Taylor expansion, the r^th term is given as: (f^(k)(a)/k!)(x-a)^k

So, if the Taylor series is truncated at the n^th term, the difference between the original function, and the n^th order Taylor polynomial, would be

(n+1)^th term + (n+2)^th term + (n+3)^th term ....

Now, for f(x)=ln(x), at a=1,

(n+1)^th term = ((-1)ⁿ/(n+1))(x-1)ⁿ⁺¹

So, f(x)-T_n(x)=(n+1)th term + (n+2)th term + ....

Since the terms are alternately positive and negative and falling progressively,

|f(c)-T_n(c)|< or = |c-1|ⁿ⁺¹/(n+1)

Now,

In |ln(1.5)-T_n(1.5)|< = 10^-2,

c=1.5

|c-1|ⁿ⁺¹/(n+1) = 10^-2

or, .5^(n+1)=0.01*(n+1)

or, n3.5

So, n=4!