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Quick question on power series of secant

  1. Jan 28, 2014 #1
    Hey, everyone.

    I am trying to find the power series of secant from the known power series of cosin, but my answer does not match up with Wolfram and Wikipedia.

    I know:
    [itex]cos(\theta) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 + ...[/itex]

    So, using the first two terms (assuming a small angle), secant should equal...

    [itex]sec(\theta) = \frac{1}{cos(\theta)} = \frac{1}{1 - \frac{1}{2}x^2} = \frac{2}{2-x^2}[/itex]

    But everywhere I go says the first two terms of the power series for secant are:

    [itex]sec(\theta) = 1 + \frac{1}{2}x^2[/itex]

    I'm sure there's something (probably elementary) that I'm missing, but I have no idea what it is. Does it have something to do with the fact that I switched the sign of the exponent when taking the reciprocal?
     
  2. jcsd
  3. Jan 28, 2014 #2

    HallsofIvy

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    You are asking for a power series, right? But [itex]\frac{2}{2- x^2}[/itex] is NOT a power series. To get the corresponding power series, divide 2 by [itex]2- x^2[/itex]. [itex]2- x^2[/itex] divides into [itex]2+ 0x+ 0x^2+ 0x^3+ \cdot\cdot\cdot[/itex] once and then [itex](2+ 0x+ 0x^2+ 0x^3+ \cdot\cdot\cdot)- (2- x^2)= x^2+ 0x^3+ \cdot\cdot\cdot[/itex]. [itex]2- x^2[/itex] divides into [itex]x^2+ 0x^3+ \cdot\cdot\cdot[itex](1/2)x^2[/itex] times and then [itex]x^2+ 0x^3+ \cdot\cdot\cdot- (1/2)x^2(2- x^2)= (1/2)x^4+\cdot\cdot\cdot[/itex] so to second power, the fraction is [itex]1+ (1/2)x^2[/itex]
     
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