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Thread: Taylor's Formula Question?

  1. #1
    Junior Member
    Jan 2010

    Taylor's Formula Question?

    Hello, I know how to rederive & prove Taylor's Formula as being a simple application
    of Integration By Parts to ∫f'(x) dx = f(b) - f(a) leading to

    f(b) = f(a) + f'(a)(b - a) + f''(a)(b - a) + ... + Rⁿ

    and that we can approximate f(x) by setting a = 0 & b = x

    f(x) = f(0) + f'(0)x + f''(0)x + ... + Rⁿ

    and I understand that Rⁿ lies between the lower & upper bound, i.e.
    ∃ c ∈ (a,b) : f ⁿ (c)(b - a)ⁿ/n! = Rⁿ
    & f ⁿ (c)(b - a)ⁿ/n! = Rⁿ

    but I'm confused on how to apply all of this.

    For example, f(x) = sinx

    okay, I can create the taylor polynomial

    sinx = x - x/3! + x^5/5! - ... + Rⁿ

    and the question is compute sin(0.1) to 3 decimals.

    a = 0,
    x = 0 + 0.1 = 0.1

    If I compute it to 3 decimals I get

    sin(0.1) = (0.1) - (0.1)/3! + (0.1)^5/5! = 0.099...

    Would this be the 3 decimals? Like, this is not correct at all compared to my calculator's answer for sin(0.1).
    the way the book has done it has me really confused.
    When I add the error on I get no significant change,
    I'm confused as to what the point of the formula is here.

    I used my calculator for this btw (isn't this equation supposed to do away with calculators? ) ?

    Now, my book has done something very different & I don't understand.
    (Serge Lang: A First Course in Calculus - page 309)

    Lang ignores calculating the Rⁿ term, and hasn't shown me how to get it yet, but
    uses the sin/cosine natural upper bound of 1, so Rⁿ ≤ Mⁿ|xⁿ|/n!
    where Mⁿ = 1 as an upper bound.

    (As a side note, this comes from the proof that ∃ c ∈ (a,b) : f ⁿ (c)(b - a)ⁿ/n!
    where he's chosen f ⁿ (v)(b - a)ⁿ/n! = Mⁿ = 1, i.e. v being the maximum,
    is the logic correct or am I missing something? He's just squeezing in
    Rⁿ underneath)

    With this he just calculates
    Rⁿ ≤ Mⁿ|xⁿ|/n! -----------> Rⁿ ≤ 1|(0.1)|/3!
    Rⁿ ≤ (0.001)/6

    Now, he says that "such accuracy would put us in the required range of accuracy.
    Hence we can just use the first term of Taylor's formula."

    sin(0.1) = 0.100 (0.001)/6

    I have no idea how or why he just did all that, non whatsoever...

    I don't understand how I could get an incorrect answer by using the x/3! & x/5!
    because the added values are supposed to give a better approximation, not a
    worse one
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  2. #2
    MHF Contributor

    Apr 2005
    Set your calculator to "Radian mode"! sin(.1) is, in fact, 0.099...
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