 #1
 141
 2
 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<(ba)^3/12n^2max(f''(x))
E<(ba)^5/180n^4max(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?
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?
Attachments
Last edited: