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Could someone help me get my head around this Taylor Series stuff

  1. May 27, 2008 #1
    1. The problem statement, all variables and given/known data

    The Taylor series for f(x) = ln(sec(x)) at a = 0 is sum_(n=0to infinity) c(sub n) (x)^n.

    Find the first few coefficients.

    3. The attempt at a solution

    I've been trying to figure out where to start by looking it up...I've seen instructions that each coefficient is just the nth derivative of the function, other's saying it's the nth derivative divided by n!, others say it's the nth derivative times n...and nothing comes out properly.

    Help!
     
  2. jcsd
  3. May 27, 2008 #2

    rock.freak667

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    The Taylor series of a function,f(x) about a point x=a is given by:

    [tex]f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+\frac{f'''(a)(x-a)^3}{3!}+\frac{f''''(a)(x-a)^4}{4!}+...+\frac{f^n(a)(x-a)^n}{n!}+...[/tex]


    as you can see, the nth term of a Taylor series is [itex]\frac{f^n(a)(x-a)^n}{n!}[/itex].
    But what you need to find are the first few terms, so find f(a), then f'(a) and so forth and just put those values into the equation above.
     
  4. May 27, 2008 #3
    Okay, so I did...

    ln(sec(0)) = 0, first term, this is correct...

    (sec(1) * tan(1) / sec(1)) * (1 - 0) = 1.5574, but the correct answer is 0...ugh!

    Help?
     
  5. May 27, 2008 #4

    rock.freak667

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    Your f(x)=ln(secx). You chose to expand about the point x=0. So the 'a' in the equation is 0

    f'(x)=secxtanx/secx, f'(0)=sec0tan0/sec0=0.
     
  6. May 27, 2008 #5
    Is (sec(x)*(sec(x)*sec(x)^2+tan(x)*sec(x)*tan(x))-sec(x)*tan(x)*sec(x)*tan(x))/sec(x)^2 f''(x)?

    EDIT: Jeez, these derivatives get huge...I must be working too hard somehow.
     
    Last edited: May 27, 2008
  7. May 27, 2008 #6

    Defennder

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    No it does not.

    f'(x) is given above by rock.freak667. Just differentiate it to get the answer. Note that you can cancel out sec x in the given expression to make your life much, much easier.
     
  8. May 28, 2008 #7

    Dick

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    Yes, it is right. If you simplify that godawful mess by cancelling sec's, you just get sec^2(x). You are right. But listen to Defennder, you are working WAY TOO HARD.
     
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