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

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SUMMARY

The Taylor polynomial of the natural logarithm function, ln(x), about the point x=1 alternates in sign due to the nature of its derivatives. Specifically, the nth derivative of ln(x) evaluated at x=1 results in alternating values: -n! for even n and n! for odd n. Consequently, the coefficients of the Taylor series expansion are 1 for odd n and -1 for even n, leading to the series representation of ln(x) as (x-1) - (1/2)(x-1)^2 + (1/3)(x-1)^3 - ... . This behavior is a direct result of the properties of the derivatives of ln(x) and their evaluation at the point of expansion.

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

I cannot grasp the concept..
 
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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.
 

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