Question

Why Taylor Series for ln(1+x) at x=0 does not equal the Taylor Series for ln(x) at x=1 ?

Answer 1

Taylor Series for ln(1+x) at x=0 vs ln(x) at x=1 . \[P^n_{c}\text{ is the n-th degree Taylor polinomial at c}\\P^3_0(\ln(1+x)) \ne P^3_1 (\ln(x)) \\ (x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3} (x-1)^3\ne x-\frac{x^2}{2}+\frac{x^3}{3} \] What is the reason for the difference? (other than because the formula says so)

Answer 2

let x = 1 + u
ln(x) = ln(1+u) at u = 0
except that your variable is u = (x - 1) the coefficients are same ... not the variables or interval

Answer 3

thanks dude, I just felt dumb right now

Answer 4

no matter where you expand it ... avoiding singularity, you will get the same value after certain terms, though some are easier ... for eg, ln(2) series representation is easier on x=0 while (1-0) = 1 while for ln(3) series representation, x=1 is easier because ln(1 + 2) => interval is (2-1) = 1