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Math Help - Stratergies for tricky Taylor Expansions

  1. #1
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    Stratergies for tricky Taylor Expansions

    I am required to taylor expand ln(sec(x)) to obtain the first 3 non-zero terms. What I wanted to know was if anyone knows a quicker method than differentating 6 times (this gets very messy)?

    Thanks in advance.
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    Note that \displaystyle \int{\tan{x}\,dx} = \ln{|\sec{x}|}.

    So find the Taylor series for \displaystyle \tan{x}, and integrate it term by term to find the Taylor series for \displaystyle \ln{|\sec{x}|}.
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    Ignore this. I didn't realise that only the first three terms were wanted.
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    Quote Originally Posted by Resilient View Post
    Isn't that very hard to find?
    Not particularly. It's usually given in a table or can be found quite easily on Wikipedia...
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    Quote Originally Posted by Prove It View Post
    Note that \displaystyle \int{\tan{x}\,dx} = \ln{|\sec{x}|}.

    So find the Taylor series for \displaystyle \tan{x}, and integrate it term by term to find the Taylor series for \displaystyle \ln{|\sec{x}|}.
    Thanks, that makes sense. Another part of the question asked me to expand tan(x+pi/4) for the first three terms. I did this by the following method:

    tan(x+pi/4) cos(x + pi/4) = sin(x + pi/4)
    tan(x+pi/4) * 0.5 *root(2) (cos(x) - sin(x)) = 0.5 * root(2) (sin(x) + cos(x))
    tan(x+pi/4) (cos(x) - sin(x)) = sin(x) + cos(x)

    I then let tan(x+pi/4) = a + bx + cx^2 , subbed in the expansions for sin(x) and cos(x) and equated coefficients. Again, can you see an improvement upon this method? Time is very precious in the exam so I'm trying to optimise my working.
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    Why don't you just substitute \displaystyle x + \frac{\pi}{4} into the series for \displaystyle \tan{X}?
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    Quote Originally Posted by Prove It View Post
    Why don't you just substitute \displaystyle x + \frac{\pi}{4} into the series for \displaystyle \tan{X}?
    I get the following working trying to follow this method which seems to be much more difficult:

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