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

  • Thread starter The Head
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  • #1
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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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Answers and Replies

  • #2
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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)|.
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.
 
  • #3
141
2
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
35,134
6,880
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.
 

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