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Math Help - series problem

  1. #1
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    series problem

    Use the Maclaurin series to make the S4 approximation of sin 3. What do you know about the error of this approximation?

    The Maclaurin series is of sin(x) in the form of x - (x^3)/3! + (x^5)/5! .................

    My book says the actual value of sin 3 lies in the interval of (0.0910714286, 0.1453125) but I don't understand why it shouldn't be (0.0368303572, 0.1453125) because a5 (which is the error bound) equals 0.0542410714. Why didn't the book include the left side of the error bound while some answers are that way??? I don't get it.

    I'm begging you to help me, my book does a terrible job explaining approximation topics. You're my only hope.
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  2. #2
    Senior Member
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    I'm not sure what you mean by S_4. Is it

    3-\frac{3^3}{3!}+\frac{3^5}{5!}-\frac{3^7}{7!}?

    I know that the remainder of the Maclaurin series is given by

    <br />
\begin{aligned}<br />
R_n(x)&=\frac{1}{(n-1)!}\int_0^xf^{(n)}(t)(x-t)^{n-1}\,dt\\<br />
&=\frac{f^{(n)}(c_n)}{n!}x^n\;\;\;\;\;\;\;\;\;\;0\  le c_n\le x.<br />
\end{aligned}<br />

    In this case, it would be

    R_8(3)=\frac{\sin c_8}{8!}3^8\;\;\;\;\;\;\;\;\;\;0\le c_n\le 3.
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  3. #3
    MHF Contributor matheagle's Avatar
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    This is an alternating series, so you can use \bigl|S_n-L\bigr|<a_{n+1}.
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