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Taylor Polynomial of Ln (x)

  1. Nov 17, 2013 #1
    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. jcsd
  3. Nov 17, 2013 #2

    HallsofIvy

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    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.
     
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