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Math Help - Taylor Series Problem

  1. #1
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    Taylor Series Problem

    Use the formula for a taylor series of a function f at c to compute the first 3 non-sero terms and the general term of the Taylor series for f(x)=sin2x at c=pi/4. Then write the taylor series for f(x)=sin2x at c=pi/4.

    Can someone explain to me how I go about solving this?
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  2. #2
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    Re: Taylor Series Problem

    A terms of the Taylor Series \sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n are those \frac{f^{(n)}(c)}{n!}(x-c)^n that are being added.

    The "general term" is \frac{f^{(n)}(c)}{n!}(x-c)^n, so say exactly what that is for this f and c, in terms of n.

    The "first three terms" are:

    \frac{ f^{(0)} (c) }{ 0! } (x-c)^0 = f(c) (but numerically computed from f and c),

    \frac{ f^{(1)} (c) } { 1! } (x-c)^1 = f'(c)(x-c) (but numerically computed from f and c),

    \frac{ f^{(2)} (c) } {2!} (x-c)^2 = \frac{f''(c)}{2}(x-c)^2 (but numerically computed from f and c).

    However, those might not be the first three non-zero terms.

    The answer wants those f and c references replaced with values as determined by what's given for f and c.

    For instance, the first term is: f(c) = \sin(2c) = \sin\left(2\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{\pi}{2}\right) = 1.

    The second term is: f'(c)(x-c) = [2\cos(2c)](x-c) = \left[ 2\cos \left( 2 \left( \frac{\pi}{4} \right) \right) \right] \left( x - \frac{\pi}{4}\right)

    = \left[2\cos\left(\frac{\pi}{2}\right)\right] \left(x - \frac{\pi}{4}\right) = 0. (Notice that this is NOT a "non-zero term").

    The 3rd term is: \frac{f''(c)}{2}(x-c)^2 = \left[ \frac{-4\sin(2c)}{2} \right] (x-c)^2 = -2 \sin \left( 2 \left( \frac{\pi}{4} \right) \right) \left( x - \frac{\pi}{4}\right)^2

    = -2 \sin \left(\frac{\pi}{2}\right) \left(x - \frac{\pi}{4}\right)^2 = -2(1)\left(x - \frac{\pi}{4}\right)^2 = -2\left(x - \frac{\pi}{4}\right)^2.
    Last edited by johnsomeone; October 25th 2012 at 05:37 PM.
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  3. #3
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    Re: Taylor Series Problem

    Use the definition of the taylor series.Compute the derivatives of sin2x at pi/4 until you have 3 nonzero derivativs.Try to figure out the general form and write the general element of the series.
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  4. #4
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    Re: Taylor Series Problem

    Quote Originally Posted by johnsomeone View Post
    A "term" of the Taylor Series \sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n are those \frac{f^{(n)}(c)}{n!}(x-c)^n that are being added.

    The "general term" is \frac{f^{(n)}(c)}{n!}(x-c)^n, so say exactly what that is for this f and c, in terms of n.

    The "first three terms" are:

    \frac{ f^{(0)} (c) }{ 0! } (x-c)^0 = f(c) (but numerically computed from f and c),

    \frac{ f^{(1)} (c) } { 1! } (x-c)^1 = f'(c)(x-c) (but numerically computed from f and c),

    \frac{ f^{(2)} (c) } {2!} (x-c)^2 = \frac{f''(c)}{2}(x-c)^2 (but numerically computed from f and c).

    However, those might not be the first three non-zero terms.

    The answer wants those f and c references replaced with values as determined by what's given for f and c.

    For instance, the first term is: f(c) = \sin(2c) = \sin\left(2\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{\pi}{2}\right) = 1.

    The second term is: f'(c)(x-c) = [2\cos(2c)](x-c) = \left[ 2\cos \left( 2 \left( \frac{\pi}{4} \right) \right) \right] \left( x - \frac{\pi}{4}\right)

    = \left[2\cos\left(\frac{\pi}{2}\right)\right] \left(x - \frac{\pi}{4}\right) = 0.

    The 3rd term is: \frac{f''(c)}{2}(x-c)^2 = \left[ \frac{-4\sin(2c)}{2} \right] (x-c)^2 = -2 \sin \left( 2 \left( \frac{\pi}{4} \right) \right) \left( x - \frac{\pi}{4}\right)^2

    = -2 \sin \left(\frac{\pi}{2}\right) \left(x - \frac{\pi}{4}\right)^2 = -2(1)\left(x - \frac{\pi}{4}\right)^2 = = -2\left(x - \frac{\pi}{4}\right)^2.
    When it says "non-zero terms" is it wanting (f^n(c)/n!)(x-c)^n when it doesn't equal 0 when you plug in f and c?
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  5. #5
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    Re: Taylor Series Problem

    Correct. It's the first 3 terms (starting with n=0), such that when you plug in for f and c, that the term isn't zero. I've already done the first two for you.

    The first two non-zero terms are: 1 \text{ and } -2\left(x - \frac{\pi}{4} \right)^2.

    You now need only find the 3rd non-zero term.
    Thanks from Preston019
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  6. #6
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    Re: Taylor Series Problem

    just the coefficient.(x-c)^n is clearly 0 at c.
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