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Math Help - Terms of sec(z) series

  1. #1
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    Terms of sec(z) series

    The question
    Find the first three non-zero terms of the Taylor series for sec(z) about 0.

    My attempt
    f(z) = \frac{1}{cos(z)}
    Radius of convergence depends on the zeros of cos(z), so |z| < \frac{\pi}{2}

    Now, I'm not sure how to proceed. I know the series for cos(z), but I'm not sure how to use it in this context. Any advice?
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  2. #2
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    Re: Terms of sec(z) series

    \displaystyle \frac{1}{\cos{(z)}} = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots .

    At \displaystyle z = 0, \frac{1}{\cos{(0)}} = c_0 + c_1(0) + c_2(0)^2 + c_3(0)^3 + \dots \implies c_0 = 1.

    Differentiate both sides

    \displaystyle \frac{\sin{(z)}}{\cos^2{(z)}} = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots

    At \displaystyle z = 0, \frac{\sin{(0)}}{\cos^2{(0)}} = c_1 + 2c_2(0) + 3c_3(0)^2 + 4c_4(0)^3 + \dots \implies c_1 = 0.

    Differentiate both sides

    \displaystyle \frac{\cos^2{(z)} + 2\sin{(z)}}{\cos^3{(z)}} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + \dots

    At \displaystyle z = 0, \frac{\cos^2{(0)} + 2\sin{(0)}}{\cos^3{(0)}} = 2c_2 + 3\cdot 2c_3(0) + 4\cdot 3c_4(0)^2 + 5\cdot 4c_5(0)^3 + \dots \implies c_2 = \frac{1}{2} .

    Keep going like this until you have the required number of terms.
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  3. #3
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    Re: Terms of sec(z) series

    How did you get that first line?
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  4. #4
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    Re: Terms of sec(z) series

    You have to write your function as an arbitrary polynomial centred about a point, in this case, \displaystyle z = 0.
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    Re: Terms of sec(z) series

    Ahh I see, I think I get what you're doing. Thanks.

    Say I wanted to get the series representation rather than just the coefficients, what would be the best way to go about that?
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    Re: Terms of sec(z) series

    Quote Originally Posted by Glitch View Post
    Ahh I see, I think I get what you're doing. Thanks.

    Say I wanted to get the series representation rather than just the coefficients, what would be the best way to go about that?
    Getting the coefficients gives you the series representation...
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  7. #7
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    Re: Terms of sec(z) series

    Forgive my ignorance, but how would I get a general representation of the series, using summation?
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    Re: Terms of sec(z) series

    Quote Originally Posted by Glitch View Post
    Forgive my ignorance, but how would I get a general representation of the series, using summation?
    You would have to try to find a pattern in the coefficients.
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  9. #9
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    Re: Terms of sec(z) series

    Thanks. That's the reason I asked, by the way, just in case there was a better way. Time is of the essence in an exam, after all.
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  10. #10
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    Re: Terms of sec(z) series

    Quote Originally Posted by Glitch View Post
    Thanks. That's the reason I asked, by the way, just in case there was a better way. Time is of the essence in an exam, after all.
    In an exam they would either give you the series, or they would ask you to find a reasonably easy series.
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