Estimating Error w/ Trap and Simpson's Rule when Values are Equal

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But the problem is hinting that there is a better way to combine the two techniques. A "best estimate" will use both methods and take advantage of both error termss.In summary, the problem involves finding estimates of a periodic function using its graph and calculating the area under one period using Simpson's Rule and the Trapezoid Rule. Part (d) of the problem asks for the best estimate of the difference between the Trapezoid estimate and the actual integral, and hints at a formula for the error in using the Trapezoid Rule. The question of whether y2 is assumed to be positive or negative is resolved by looking at the graph. It is also noted that Simpson's Rule and the Trapezoid Rule have different error
  • #1
The Head
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Homework Statement
I'm attaching the full problem with graph, but the problem has me find estimates of a periodic function using its graph by finding the area of a full period by using:
a) Simpson's Rule, starting at two different locations in the periodic function
b) Finding the average of these two estimates
c) Using the Trapezoidal Rule to estimate the area under one period
These are all very straight forward.

Then, part (d) has me suspicious. It tells me |max(f''(x))| AND |max(f''''(x))| on the interval =1 and asks me for the BEST estimate of |Trap estimate - integral (f(x))|
Relevant Equations
E<(b-a)^3/12n^2|max(f''(x))|
E<(b-a)^5/180n^4|max(f''''(x)|
I'm confident in calculating the beginning values. Only question I had was whether in this graph y2 is assumed positive or negative. Obviously f(pi)= negative value, but is y2 negative or is it = |f(pi)|. Assuming the latter, it all makes sense and when I sum the two Simpson's Rule estimates I get:
pi/2(y_0 + y_1 - y_2 +y_3)

I get the exact same thing for the trapezoidal rule. All of this makes sense. But the wording of "best estimate" in part (d) of |Trap estimate - integral (f(x))| has me suspicious that it's not just the error formula.

Further, I figure that since an average of the Simpson's Errors gives me a smaller value, the error for the trapezoidal estimate (since it's the same as the average of Simpson's) is less than this as well. This feels like a bit of an assumption, but I guess I can say:

T= 1/2(S1+S2), so there exists some estimate S3 that equals T (by intermediate value theorem). Thus, the error < that calculated by the Simpson's Rule estimate. Is that about as far as I can go?
 

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  • #2
The Head said:
Homework Statement:: I'm attaching the full problem with graph, but the problem has me find estimates of a periodic function using its graph by finding the area of a full period by using:
a) Simpson's Rule, starting at two different locations in the periodic function
b) Finding the average of these two estimates
c) Using the Trapezoidal Rule to estimate the area under one period
These are all very straight forward.

Then, part (d) has me suspicious. It tells me |max(f''(x))| AND |max(f''''(x))| on the interval =1 and asks me for the BEST estimate of |Trap estimate - integral (f(x))|
Relevant Equations:: E<(b-a)^3/12n^2|max(f''(x))|
E<(b-a)^5/180n^4|max(f''''(x)|

I'm confident in calculating the beginning values. Only question I had was whether in this graph y2 is assumed positive or negative.
Obviously f(pi)= negative value, but is y2 negative or is it = |f(pi)|.
It should be obvious from the graph that y2 is negative.
The Head said:
Assuming the latter, it all makes sense and when I sum the two Simpson's Rule estimates I get:
pi/2(y_0 + y_1 - y_2 +y_3)

I get the exact same thing for the trapezoidal rule. All of this makes sense. But the wording of "best estimate" in part (d) of |Trap estimate - integral (f(x))| has me suspicious that it's not just the error formula.

Further, I figure that since an average of the Simpson's Errors gives me a smaller value, the error for the trapezoidal estimate (since it's the same as the average of Simpson's) is less than this as well. This feels like a bit of an assumption, but I guess I can say:

T= 1/2(S1+S2), so there exists some estimate S3 that equals T (by intermediate value theorem). Thus, the error < that calculated by the Simpson's Rule estimate. Is that about as far as I can go?
I think that what the problem is hinting at is a formula for the error in using the Trapezoid Rule. Has your textbook presented any discussion about the error bounds for this method? Here's some more information: https://en.wikipedia.org/wiki/Trapezoidal_rule, in the section titled Error analysis.
 
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Mark44 said:
It should be obvious from the graph that y2 is negative.
I think that what the problem is hinting at is a formula for the error in using the Trapezoid Rule. Has your textbook presented any discussion about the error bounds for this method? Here's some more information: https://en.wikipedia.org/wiki/Trapezoidal_rule, in the section titled Error analysis.
But isn't y_2 showing a length? And why would they spend so much time going through Simpson's if that isn't applied here? It seems fishy they would give f''''(x), point to the idea that the two estimates are the same, and ask for "best estimate." Plus since the average of Simpson = the Trapezoidal, if they have the same estimate, then wouldn't both be within the same error rate of Simpson (which is smaller than the Trapezoidal).
 
  • #4
The Head said:
But isn't y_2 showing a length?
##y_0, y_1, y_2,## and ##y_3## are coordinates, so can be positive or negative or zero. If your formula is using a positive value for ##y_2## your result is wrong.
The Head said:
And why would they spend so much time going through Simpson's if that isn't applied here? It seems fishy they would give f''''(x), point to the idea that the two estimates are the same, and ask for "best estimate." Plus since the average of Simpson = the Trapezoidal, if they have the same estimate, then wouldn't both be within the same error rate of Simpson (which is smaller than the Trapezoidal).
Simpson's Rule and the Trapezoid Rule have different error terms, one involving the 2nd derivative and the other involving the fourth derivative. For the particular function in this problem, it could well be that one technique will give a smaller error than the other, given how few subintervals are being used.
 

1. What is the purpose of estimating error with Trap and Simpson's Rule?

The purpose of estimating error with Trap and Simpson's Rule is to determine the accuracy of numerical integration methods. These methods are used to approximate the area under a curve and can be prone to error, so estimating error allows us to assess the reliability of our results.

2. How do Trap and Simpson's Rule estimate error when values are equal?

When values are equal, Trap and Simpson's Rule use different approaches to estimate error. Trap Rule uses the difference between the maximum and minimum values of the function to calculate the error, while Simpson's Rule uses the difference between the average value and the actual value of the function. Both methods use these differences to determine the accuracy of the approximation.

3. Can Trap and Simpson's Rule accurately estimate error for all types of functions?

No, Trap and Simpson's Rule are most accurate when used on continuous functions. They may not give accurate estimates for functions with discontinuities or sharp changes in the slope.

4. How can we use the estimated error to improve our results?

The estimated error can help us determine the number of intervals or subintervals needed for a more accurate approximation. By decreasing the interval size, we can reduce the error and improve the accuracy of our results.

5. Are there any limitations to using Trap and Simpson's Rule for estimating error?

One limitation of Trap and Simpson's Rule is that they assume the function is smooth and continuous, which may not always be the case. Additionally, these methods may not be suitable for highly complex functions with multiple variables. It is important to consider the limitations and potential sources of error when using these methods for estimating error.

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