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Math Help - Taylor's theorem

  1. #1
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    Taylor's theorem

    multiple choice
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  2. #2
    Grand Panjandrum
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    (1) Find the sum:

    <br />
S = x + \frac{x^2}{2\cdot 2}+ \frac{x^3}{3\cdot 2^2}+ \frac{x^4}{4\cdot 2^3} + ...<br />

    Consider this as a function of x (note the negative sign in front of the term in x^4
    I presume is a mistake), the derivative of this is:

    <br />
U(x) = 1 + \frac{x}{ 2}+ \frac{x^2}{ 2^2}+ \frac{x^3}{ 2^3} + ...<br />

    which is a geometric series and it converges for |x|<2.

    So write down the sum of this last series, then integrate with the constant
    of integration set so that the integral is zero for x=0, and that
    is your answer (it looks like (d) to me but you will need to check).

    RonL
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  3. #3
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    There are many verions, I use the the following one.

    Taylor's Theorem (Lagrange): Let f be n differenciable on (a,b) (with a<0<b) and for 0\not =x\in (a,b) the remainder is given by R_n(x) = \frac{f^{(n)}(y)}{n!}\cdot x^n for some y between 0 and x. Where the remainder is the difference between the function and its n-1 degree Taylor polynomial T_{n-1}(x), defined as T_{n-1}(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!}\cdot x^k and R_n(x) = f(x) - T_{n-1}(x).

    So, if n=5 then T_4(x) = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!} on (-.1,.1)

    Then, R_5(x) = \frac{f^{(5)}(y)}{5!}\cdot x^5 = \frac{e^y}{5!} \leq \frac{e^{.1}}{5!}\approx .003

    So it is 2 decimal points.
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