Why Does the Taylor Polynomial of Ln(x) Alternate in Sign?

[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..

 

[HallsofIvy]

The nth Taylor polynomial of f(x), about x= a is
[tex]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[/tex]
( [itex]f^{(n)}(a)[/itex] 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. [itex]d(ln(x))/dx= 1/x[/itex] which is 1 at x= 1. [itex]d^2(ln(x))/dx= d(1/x)/dx= d(x^{-1})/dx= -1/x^2[/itex] which is -1 at x= 1. Differentiating again, the derivative of [itex]-1/x^2= -x^{-2}[/itex] is [itex]2x^{-3}[/itex] which is 2 at x= 1.
Differentiating again, the derivative of [itex]2x^{-3}[/itex] is [itex]-6x^{-4}[/itex] 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 [itex](x- 1)^n[/itex] is [itex]n!/n!= 1[/itex] if n is odd, [itex]-n!/n!= -1[/itex] if n is even.