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Math Help - Taylor Series Problems

  1. #1
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    Question Taylor Series Problems

    Hi all,

    These have been giving me some trouble:

    1. Use series to approximate the definite integral to within 3 decimal places

    The integral from 0 to 1 : xcos(x^3) dx

    2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.

    a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2

    b. f(x)= e^(x^2), a=0, n=3, 0 < or equal to x < or equal to 0.1

    Thanks a lot for any help

    -coolio
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by coolio View Post
    Hi all,

    These have been giving me some trouble:

    1. Use series to approximate the definite integral to within 3 decimal places

    The integral from 0 to 1 : xcos(x^3) dx

    2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.

    a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2

    b. f(x)= e^(x^2), a=0, n=3, 0 < or equal to x < or equal to 0.1

    Thanks a lot for any help

    -coolio
    1. \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}...therefore \cos(x^3)=\sum+{n=0}^{\infty}\frac{(-1)^{n}(x^3)^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{  (-1)^{n}x^{6n}}{(2n)!}

    Now multiplying by x we get x\cos(x^3)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{6n+1}}{(2n)!}

    Integrating we get \int{x\cos(x^3)}dx=\int\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{6n+1}}{(2n)!}dx=\sum_{n=0}^{\infty}\frac{  (-1)^{n}x^{6n+2}}{(6n+2)(2n)!} and to get [tex] \int_0^{1}x\cos(x^3)dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{6n+2}}{(6n+2)(2n)!}\bigg|_0^{1}=.440
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by coolio View Post
    Hi all,

    These have been giving me some trouble:

    1. Use series to approximate the definite integral to within 3 decimal places

    The integral from 0 to 1 : xcos(x^3) dx

    2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.

    a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2

    b. f(x)= e^(x^2), a=0, n=3, 0 < or equal to x < or equal to 0.1

    Thanks a lot for any help

    -coolio
    For the other two use f(x)\approx{f(c)+(x-c)f'(c)+frac{f''(c)(x-c)^2}{2!}+\frac{(x-c)^3f'''(c)}{3!}} +...+\frac{(x-c)^nf^{(n)}}{n!}
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    For the other two use f(x)\approx{f(c)+(x-c)f'(c)+frac{f''(c)(x-c)^2}{2!}+\frac{(x-c)^3f'''(c)}{3!}} +...+\frac{(x-c)^nf^{(n)}}{n!}
    f(x)=\lim_{n\to\infty}\sum_{n=c}\frac{(x-c)^{n}f^{(n)}}{n!}

    Where f^{(n)} denotes \frac{dy^{n}}{d^{n}x}
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  5. #5
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    Quote Originally Posted by coolio View Post
    1. Use series to approximate the definite integral to within 3 decimal places

    The integral from 0 to 1 : xcos(x^3) dx
    Note x\cos x^3 = \sum_{n=0}^{\infty}(-1)^n\frac{x^{6n+1}}{(2n)!}. This means, \int_0^1 x\cos x^3 dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \int_0^1 x^{6n+1} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(6n+2)}. Let A_n be the n term approximation. Then since this series is an alternating series it means \left| \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(6n+2)} - A_n\right| \leq \frac{1}{(2n)!(6n+2)}. To approximate this to three decimal places means we want to make \frac{1}{(2n)!(6n+2)} \leq \frac{1}{1000} which happens first at n=2.
    ---
    Mathstud do not make posts which do not answer the question, the question was not asking for an infinite series, it was asking for an approximation.
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  6. #6
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    Quote Originally Posted by coolio View Post
    2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.
    a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2
    Let T_2 be the 2nd degree Taylor approximating polynomial. If x is a chosen point on (4,4.2) then it means \left|\sqrt{x} - T_2(x)\right|  = \frac{|f'''(y)|}{3!}|x-4|^3 \leq \frac{1}{6}\cdot \frac{1}{5}\cdot |f'''(y)|. Now put a bound on f'''(y).
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