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    Does anybody know how to use a first degree Taylor polynomial to approximate the value of tan(3)?

    I know the Taylor series of tan(x), but confused as to how to use it to approximate tan(3).

    Cheers.
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    Wouldn't you just substitute x = 3 and add the first few terms, until you decide that the contribution of each sucessive term is too small to make a noticable difference? Have you been given the number of places that you need your approximation to?
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    The first three terms of tan(x) are x + 1/3(x)^3 + 2/15(x)^5 I know this much, but if you substitute in 3 for x you won't get an approximation for tan(3) so obviously I am doing something wrong or have not understood something.
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    (Original post by IceCityRabz)
    The first three terms of tan(x) are x + 1/3(x)^3 + 2/15(x)^5 I know this much, but if you substitute in 3 for x you won't get an approximation for tan(3) so obviously I am doing something wrong or have not understood something.
    Why do you think that putting x=3 won't give you an approximation to tan(3)? Genuine question, I'm not sure of this - it there a range of permissible values that we have stepped outside of? If so, can we use the idea that;

    tan(3) = tan(3-pi) = -tan(pi-3),

    so sub in pi-3 and then take the negative of what you get.
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    (Original post by Pangol)
    Why do you think that putting x=3 won't give you an approximation to tan(3)? Genuine question, I'm not sure of this - it there a range of permissible values that we have stepped outside of? If so, can we use the idea that;

    tan(3) = tan(3-pi) = -tan(pi-3),

    so sub in pi-3 and then take the negative of what you get.
    Yes I think you are right. thanks
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    (Original post by Pangol)
    Wouldn't you just substitute x = 3 and add the first few terms, until you decide that the contribution of each sucessive term is too small to make a noticable difference? Have you been given the number of places that you need your approximation to?
    First degree implies you're only supposed to use the first 2 terms (i.e. f(a) + (x-a)f'(a)).

    Your other suggestion to use tan(3) = -tan(pi-3) is a good one, however.
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    (Original post by DFranklin)
    First degree implies you're only supposed to use the first 2 terms (i.e. f(a) + (x-a)f'(a)).

    Your other suggestion to use tan(3) = -tan(pi-3) is a good one, however.
    My eyes rather skipped over the "first degree" part - point taken. Out of interest, what values does the Taylor series coverge for? Is it 0 <- x < pi/2 ?
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    (Original post by Pangol)
    My eyes rather skipped over the "first degree" part - point taken. Out of interest, what values does the Taylor series coverge for? Is it 0 <- x < pi/2 ?
    Taylor series around x =0 should converge for -pi/2 < x < pi/2 I think.

    In general, given a point a in the complex plane and a function f analytic near a, if b is the nearest pole to a (you can think of a pole as a point where f goes infinite), then the radius of convergence of the Taylor series will be |b-a|.

    In this case the nearest poles are where cos x = 0, that is x=+/- pi/2.
 
 
 
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