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- Feb 28th 2010, 07:45 AMharoldSimpson's Rule
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- Feb 28th 2010, 08:18 AMCaptainBlack
- Feb 28th 2010, 01:25 PMchisigma
From my point of view there are two points that need to be better specified...

a) on how many points of the interval $\displaystyle [0,2\pi]$ the Simpson's formula has to be used?...

b) what does exactly mean 'actual error'?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 28th 2010, 03:43 PMharold
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- Feb 28th 2010, 08:04 PMCaptainBlack
The parametric equation tell you that:

$\displaystyle x(t)=t \cos(t)$

$\displaystyle y(t)=t \sin(t)$

Quote:

Recall from calculus that the arclength formula of a parametric curve $\displaystyle c(t) = (x(t), y(t))$ with $\displaystyle a\le t\le b$ is given by $\displaystyle l(c)=\int_a^b\sqrt{x'(t)^2 + y'(t)^2} \, dt$.

$\displaystyle y'(t)=\sin(t) + t \cos(t)$

So the curve length is:

$\displaystyle l(c)=\int_a^b\sqrt{[\cos(t) - t\sin(t)]^2 + [\sin(t) + t \cos(t)]^2} \, dt$

Now simplify the integrand

Quote:

Use Simpson's Rule to approximate the integral. Find the error bound using the appropriate error formula, and compare it to the actual error.

How can I solve this?

CB - Feb 28th 2010, 11:23 PMharold
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- Feb 28th 2010, 11:29 PMGeneral
- Feb 28th 2010, 11:57 PMCaptainBlack
- Mar 1st 2010, 12:09 AMCaptainBlack
Put $\displaystyle f(x)=\sqrt{1+x^2}$, then:

$\displaystyle

\int_0^{2\pi}\sqrt{1+x^2}\;dx \approx \frac{h}{3}[f(0)+2f(h)+4f(2h)+...+4f((n-2)h)+2f((n-1)h)+f(nh)]

$

where $\displaystyle h=2\pi/n$ where $\displaystyle n$ (even) is the number of steps.

The absolute value of the error in this approximation is less than:

$\displaystyle

\frac{h^4}{180} \max_{\xi \in [0, 2\pi]}|f''''(\xi)|

$

Now where is the problem exactly?

CB - Mar 1st 2010, 12:10 AMGeneral
- Mar 1st 2010, 01:51 AMchisigma
In my opinion the probelem 'exactly' is the follows...

a) we know the*exact*value of the integral...

$\displaystyle \int_{0}^{2\pi} \sqrt{1 + t^{2}}\cdot dt$ (1)

... because the integrand function has an elementar trascendent primitive...

b) we compute an approximate value of (1) using Simpson rule...

$\displaystyle

\int_0^{2\pi}\sqrt{1+t^2}\;dt \approx \frac{h}{3}[f(0)+2f(h)+4f(2h)+...+4f((n-2)h)+2f((n-1)h)+f(nh)]

$ (2)

... where $\displaystyle h=2\pi/n$...

c) the approximate value obtained with (2) is of course function of $\displaystyle h$ [or equivalently of $\displaystyle n$...] and the 'effective error' [the difference between the 'exact value' and the value obtained with (2)...] has to compared with the 'bound'...

$\displaystyle

\frac{h^4}{180} \max_{\xi \in [0, 2\pi]}|f''''(\xi)|

$ (3)

... all that for different value of $\displaystyle n$ in order to valuate the 'accuracy' as function of $\displaystyle h$... very little and pleasant job (Wink) !...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 1st 2010, 01:59 AMCaptainBlack
Also my mistake as I did not see the last clause in the problem statement where the error bound is to be compared with the actual error.

But still everything is just book work from the point of arriving at the integral of $\displaystyle \sqrt{1+x^2}$ onward (the integral can be looked up in a table of integrals if need be).

CB - Mar 1st 2010, 02:41 AMharold
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- Mar 1st 2010, 03:09 AMCaptainBlack
- Mar 1st 2010, 03:29 AMharold
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