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Math Help - Tayor Series

  1. #1
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    Tayor Series

    I need to find the gerneral term for the Taylor series expansion with x=2 and f(x) = ln(abs(x-1)).

    Not really sure where to start. Instructions say to refer to a a similar questions where f(x) = 1/(x-1). And I got E(0, infinity) (a)(2-a)^n/(n!). Not confident that's right though. I notice that 1/(1-x) is the derivative of ln(x-1), but my brain cannot quite connect the dots any further.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Nerd View Post
    I need to find the gerneral term for the Taylor series expansion with x=2 and f(x) = ln(abs(x-1)).

    Not really sure where to start. Instructions say to refer to a a similar questions where f(x) = 1/(x-1). And I got E(0, infinity) (a)(2-a)^n/(n!). Not confident that's right though. I notice that 1/(1-x) is the derivative of ln(x-1), but my brain cannot quite connect the dots any further.
    E(0, infinity) f^{n}(a)(2-a)^n/(n!)

    Each term in the sum is multiplied by the value of the "n'th" derivative of the function defined at a. So, the first term is the "0th" derivative, or the function defined at a: ln(abs(a-1)); the second term is the 1st derivative defined at a: 1/(a-1); the third is the 2nd derivative defiend at a: -1/(a-1)^2 ... etc.
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  3. #3
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    Scratch that original answer. My concepts were really rusty. I got confused with the meaning of a and x and how to apply the formula that's in my chicken-scratch handwriting. I think I got it now. I'll post my results later.
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  4. #4
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    Urg...Maybe I don't

    Okay let's start with the easy one

    f(x) = 1/(x-1) about x=2

    So...
    f(x)= 1(x-1)^(-1)
    f'(x) = -1(x-1)^(-2)
    f''(x) = -2*-1(x-1)^(-3)
    f'''(x) = -3*-2*-1(x-1)^(-4)

    So I got for f^n(x) = ((-1)^n)*(n!)*((x-1)^(-(n+1))
    So f^n(2) = ((-1)^n)*(n!)*((2-1)^(-(n+1)) = ((-1)^n)*(n!)

    So then I get E (((-1)^n)*n!*((x-2)^n))/n! = E ((-1)^n)*((x-2)^n) = 1-(x-2)+(x-2)^2-...etc.

    So is that right?
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  5. #5
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    No one? I reeeally need help on this soon.
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  6. #6
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    Pwetty pwease? Can someone walk me through this? That last problem and ln (x-1)? I think I'm supposed to find f^n(a), but can't figure out exactly how to go about that.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Nerd View Post
    Pwetty pwease? Can someone walk me through this? That last problem and ln (x-1)? I think I'm supposed to find f^n(a), but can't figure out exactly how to go about that.
    You did the one for 1/(x - 1) correctly. Here's how to use it to find ln|x - 1|. Remember, this is about x = 2
    Attached Thumbnails Attached Thumbnails Tayor Series-prob2.gif  
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  8. #8
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    Thanks a bunch. Despite my homework being hopelessly late, I think I starting to understand more of this (and that's the imporant part right?)

    One more (don't take my word on that. I'll probably be back)

    How would I use that to compute a number that differs from ln(3/2) by less than 0.05?

    First thing that came to me was to plug in x=5/2. Is that right at all? How could I tell if the error is less than 0.05?
    Last edited by Nerd; April 15th 2007 at 04:58 PM. Reason: Fixed some pronouns without a clear antecedent and punctuation.
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  9. #9
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    Nevermind. I got it.
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