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Using simpsons rule, how to find the error bound?

  • Thread starter Randall
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  • #1
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Homework Statement


Find the error bound using Simpsons Rule for integral of SQRT (x) dx from 1 to 9.[/B]


Homework Equations


E = (M * (b-a)^5) / (180 * n^4), where M = max value of the 4th deriv of x dx
[/B]


The Attempt at a Solution


see attached please - I can't figure out what to use for the value of M, which is supposed to be the maximum value of the 4th derivative. I graphed the 4th deriv (see attached) and the max y = 0 and the max x is infinity? I need help figuring out M please thanks.https://www.dropbox.com/s/3cn5q3n96iutgu4/IMG_6352.JPG?dl=0 [/B]
 

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  • #2
Ray Vickson
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Homework Statement


Find the error bound using Simpsons Rule for integral of SQRT (x) dx from 1 to 9.[/B]


Homework Equations


E = (M * (b-a)^5) / (180 * n^4), where M = max value of the 4th deriv of x dx
[/B]


The Attempt at a Solution


see attached please - I can't figure out what to use for the value of M, which is supposed to be the maximum value of the 4th derivative. I graphed the 4th deriv (see attached) and the max y = 0 and the max x is infinity? I need help figuring out M please thanks.https://www.dropbox.com/s/3cn5q3n96iutgu4/IMG_6352.JPG?dl=0 [/B]
M is not the max value of the fourth derivative of x dx (although it is related to the fourth derivative of ##f(x)## = the integrand in ##\int_a^b f(x) \, dx##). In fact,
[tex] M = \max_{a \leq x \leq b} |f^{(4)} (x) | [/tex]

What happens near x = 0 or out at x = ∞ is irrelevant, since the only values the matter here are the ones between x = 1 and x = 9.
 
  • #3
Zondrina
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Assuming the 1/3 rule, the approximate error is given by:

$$E_a = - \frac{(b-a)^5}{180n^4} \space \bar f^{(4)}(\epsilon)$$

Where ##\bar f^{(4)}(\epsilon) = \frac{\int_a^b f^{(4)}(x) \space dx}{b - a}## is the average fourth derivative on the interval.
 
  • #4
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M is not the max value of the fourth derivative of x dx (although it is related to the fourth derivative of ##f(x)## = the integrand in ##\int_a^b f(x) \, dx##). In fact,
[tex] M = \max_{a \leq x \leq b} |f^{(4)} (x) | [/tex]

What happens near x = 0 or out at x = ∞ is irrelevant, since the only values the matter here are the ones between x = 1 and x = 9.

Thank you!
-R
 
  • #5
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0
Assuming the 1/3 rule, the approximate error is given by:

$$E_a = - \frac{(b-a)^5}{180n^4} \space \bar f^{(4)}(\epsilon)$$

Where ##\bar f^{(4)}(\epsilon) = \frac{\int_a^b f^{(4)}(x) \space dx}{b - a}## is the average fourth derivative on the interval.

Thank you!
-R
 
  • #6
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Ok. So after calculating the error bound using M=15/16, I get 0.6667. Does this mean the approximation needs to be within 0.66% of actual? Or does it need to be within 66.67% of actual? 66.67% seems awfully big, and 0.66% seems to be awfully small....
 
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  • #7
haruspex
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Ok. So after calculating the error bound using M=15/16, I get 0.6667. Does this mean the approximation needs to be within 0.66% of actual? Or does it need to be within 66.67% of actual?
Neither. What makes you think it's a fractional or percentage error?
 
  • #8
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Oh! Ok. So what do I do with this 0.6667 value then? What does that tell me? The problem wants to verify that the approximate value (found via the Simpsons rule) of 17.32 is within the error bound (0.6667) of the actual value of 17.6667. Do I add 0.6667 to 17.32 = 18, and since the actual value is 17.6667 (which is less than 18), I am within the error bound?
 
  • #9
haruspex
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Oh! Ok. So what do I do with this 0.6667 value then? What does that tell me? The problem wants to verify that the approximate value (found via the Simpsons rule) of 17.32 is within the error bound (0.6667) of the actual value of 17.6667. Do I add 0.6667 to 17.32 = 18, and since the actual value is 17.6667 (which is less than 18), I am within the error bound?
Yes. Logically you would run that the other way: is the calculated value between (actual - bound) and (actual + bound), but it comes to the same thing.
 

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