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Taylor series (very easy but have a problem)

  1. Jun 24, 2011 #1
    1. The problem statement, all variables and given/known data

    series expansion at [tex]c=2[/tex] of [tex]ln(x^2+x-6)[/tex]

    2. Relevant equations



    3. The attempt at a solution

    After substituting [tex]y= x -2[/tex] we get ln(y^2+5y) = ln(y) + ln(y+5) but im not kinda sure how to use the taylor series of ln(1+x)...
     
  2. jcsd
  3. Jun 24, 2011 #2

    gb7nash

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    Hi Fermatpall,

    Try factoring x^2+x−6 instead. So:

    ln(x^2+x−6) = ln((x+3)(x-2)) = ln(x+3) + ln(x-2)

    The problem tells us to expand the taylor series around c = 2. The formula we need is:

    [tex]\sum_{n = 0} ^ {\infty} \frac{f^{(n)}(c)}{n!} \, (x - c)^{n}[/tex]

    Let's crank out some derivatives then:

    [tex]f'(x) = \frac{1}{x+3} + \frac{1}{x-2}[/tex]
    .
    .
    .

    Calculate a few more derivatives and you'll find a pattern. After this is done, throw your result into the formula.
     
  4. Jun 24, 2011 #3

    I like Serena

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    Hmm, I don't think there is a Taylor expansion around x=2, since [itex]\ln(2^2 + 2 -6)=-\infty[/itex].
     
  5. Jun 24, 2011 #4

    gb7nash

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    Good call. I didn't check to see if the point was even valid.
     
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