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Math Help - Numerical Analysis Midpoint Rule Help

  1. #1
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    Numerical Analysis Midpoint Rule Help

    Could someone check if this is the correct way to solve this problem.

    For what n will the approximation f(x) be accurate to within .0000001 using

    E_{f} = \frac{(b-a)h^2}{6} f''(\mu), for some \mu between a and b is.

    h= \frac{b-a}{n+2}

    I think f''(\mu) for some \mu between a and b means the absolute value of the max between a and b.

    f(x) = \int_0^1(\frac{1}{12}x^4 - \frac{1}{6}x^3)dx

    .0000001 \leq \frac{(1-0)}{6} (\frac{1-0}{n+2})^2 f''(\mu)

    f''(\mu) = \mu^2 - \mu

    .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * (\mu^2 - \mu)

    From this part on am I suppose to maximize \mid f''(\mu)\mid on [0,1] and put that in place of f''(\mu) and then solve for n?

    If I do that I get

    .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * 5, where 5 is the max of \mid f''(\mu)\mid on [0,1]

    Solving this for n gives 2886.75 and rounding it to 2887. So n would have to be 2887.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by TshingY View Post
    Could someone check if this is the correct way to solve this problem.

    For what n will the approximation f(x) be accurate to within .0000001 using

    E_{f} = \frac{(b-a)h^2}{6} f''(\mu), for some \mu between a and b is.

    h= \frac{b-a}{n+2}

    I think f''(\mu) for some \mu between a and b means the absolute value of the max between a and b.

    f(x) = \int_0^1(\frac{1}{12}x^4 - \frac{1}{6}x^3)dx

    .0000001 \leq \frac{(1-0)}{6} (\frac{1-0}{n+2})^2 f''(\mu)

    f''(\mu) = \mu^2 - \mu

    .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * (\mu^2 - \mu)

    From this part on am I suppose to maximize \mid f''(\mu)\mid on [0,1] and put that in place of f''(\mu) and then solve for n?

    If I do that I get

    .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * 5, where 5 is the max of \mid f''(\mu)\mid on [0,1]

    Solving this for n gives 2886.75 and rounding it to 2887. So n would have to be 2887.
    You have jumbled things up, please post the original question

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by TshingY View Post
    Could someone check if this is the correct way to solve this problem.

    For what n will the approximation f(x) be accurate to within .0000001 using

    E_{f} = \frac{(b-a)h^2}{6} f''(\mu), for some \mu between a and b is.

    h= \frac{b-a}{n+2}

    I think f''(\mu) for some \mu between a and b means the absolute value of the max between a and b.

    f(x) = \int_0^1(\frac{1}{12}x^4 - \frac{1}{6}x^3)dx

    .0000001 \leq \frac{(1-0)}{6} (\frac{1-0}{n+2})^2 f''(\mu)

    f''(\mu) = \mu^2 - \mu

    .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * (\mu^2 - \mu)

    From this part on am I suppose to maximize \mid f''(\mu)\mid on [0,1] and put that in place of f''(\mu) and then solve for n?

    If I do that I get

    .0000001 \leq \frac{1}{6} (\frac{1}{n+2})^2 * 5, where 5 is the max of \mid f''(\mu)\mid on [0,1]

    Solving this for n gives 2886.75 and rounding it to 2887. So n would have to be 2887.
    The maximum of g(x)=|f''(x)|=|x^2-x| on [0,1] occurs at x=1/2 and so the maximum is: g(1/2)=1/4.

    CB
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