Taylor Polynomial of Ln (x)

1. Nov 17, 2013

xtrubambinoxpr

I need help understanding why the ln (x) taylor polynomial is (x-1)-1/2(x-1)^2.... + etc.

I cannot grasp the concept..

2. Nov 17, 2013

HallsofIvy

Staff Emeritus
The nth Taylor polynomial of f(x), about x= a is
$$f(a)+ f'(a)(x- a)+ \frac{f''(a)}{2}(x- a)^2+ \frac{f'''(a)}{3!}(x- a)^3+ ...+ \frac{f^{(n)}(a)}{n!}(x- a)^n$$
( $f^{(n)}(a)$ indicates the nth derivative of f evaluated at x= a)
You've probably seen that!

For f(x)= ln(x), at a= 1 (we cannot use a= 0 because ln(0) is not defined) ln(1)= 0. $d(ln(x))/dx= 1/x$ which is 1 at x= 1. $d^2(ln(x))/dx= d(1/x)/dx= d(x^{-1})/dx= -1/x^2$ which is -1 at x= 1. Differentiating again, the derivative of $-1/x^2= -x^{-2}$ is $2x^{-3}$ which is 2 at x= 1.
Differentiating again, the derivative of $2x^{-3}$ is $-6x^{-4}$ which is -6 at x= 1. Do you see the point? The "nth" derivative of ln(x) alternates sign and is -n! for n even and n! for n odd.

That means the coefficient of $(x- 1)^n$ is $n!/n!= 1$ if n is odd, $-n!/n!= -1$ if n is even.