Numerical Analysis Midpoint Rule Help

**TshingY** · May 4th 2010, 10:44 AM

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.

**CaptainBlack** · May 5th 2010, 11:39 PM

Originally Posted by TshingY

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

**CaptainBlack** · May 5th 2010, 11:46 PM

Originally Posted by TshingY

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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