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NEED HELP! with power series! PLEASE!

  1. Aug 5, 2004 #1
    I need to know how to get the first 6 terms of the power series for:
    f(x)=ln(x) centered at x=1. Thanks
     
  2. jcsd
  3. Aug 5, 2004 #2
    Well

    [tex]f'(x)=\frac {1}{x}[/tex]
    [tex]f''(x)= - \frac {1}{x^2}[/tex]
    [tex]f'''(x)= \frac {2}{x^3}[/tex]
    [tex]f^4(x)=- \frac {6}{x^4}[/tex]
    [tex]f^5(x)=\frac {24}{x^5}[/tex]
    [tex]f^6(x)=- \frac {120}{x^6}[/tex]

    so [tex]f(1)=0, \ f'(1)=1, \ f''(1)=-1 , \ f'''(1)=2, \ f^4(1)=-6 , ...[/tex]

    so [tex]\ln x \approx 0 + (x-1) + (-1) \frac {(x-1)^2}{2} + (2) \frac {(x-1)^3}{3!} + (-6) \frac {(x-1)^4}{4!}+ (24)\frac {(x-1)^5}{5!}[/tex]
     
  4. Aug 5, 2004 #3
    thanks, and that 2nd term answer isnt supposed to be 2! on the bottom instead of 2 is it?
     
  5. Aug 5, 2004 #4

    arildno

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    That's correct (even though, of course 2!=2); the full series expansion of ln(x) about
    x=1 is
    [tex]ln(x)=\sum_{i=1}^{\infty}(-1)^{i+1}\frac{(x-1)^{i}}{i}[/tex]
     
  6. Aug 6, 2004 #5

    mathwonk

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    the usual way to do this is to note that 1/(1+x) = 1 - x + x^2 - x^3 + x^4 -+ .....

    Then integrate both sides. On the left you get ln(1+x) and on the right you get

    well you do it.... This will be valid wherever the series on the right converges, i.e. for |x| < 1. but also for x = 1, which yields a nice series for ln(2).
     
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