Error-bounds - MA104 Lecture Notes PDF

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Error bounds Remember that midpoint rule, trapezoidal rule, and Simpson’s different ways to come up with an approximation for area un But how do we know how accurate our approximation is, in co the exact area under the curve? We want to know whether an approximation is very good, and close to actual area, or if it’s approximation of actual area. That’s where the error bound formulas come in. They tell us t possible error in our approximations. So if the error bound is we know that it’s possible that our approximation is bad, and actual area. If the error bound is very small, we know that our approximation is pretty good, and close to the actual area. The error bound formulas are Midpoint rule error bound

Trapezoidal rule error bound

Simpson’s rule error bound h

EM

K(b − a)3 ≤ 24n 2

ET

K(b − a)3 ≤ 12n 2

ES

K(b − a)5 ≤ 180n 4

• EM , ET , and ES are the absolute values of the ac which you can also think of as the maximum possible maximum possible difference between your approxim and actual area • [a, b] is the interval over which you’re finding area • n is the number of subintervals you’re using to find ar interval [a, b] • f′′ ( x) is the second derivative of the given function f (x) • f (4)(x) is the fourth derivative of the given function f (x You’ll want to use information from your problem to plug in fo and you’re going to be solving the inequality for EM , ET , o means the only value you really need to find is K. Finding K is tricky part when it comes to finding error bound. Notice that for midpoint and trapezoidal rules, f′′ (x)  ≤ K, and Simpson’s rule f (4)(x) ≤ K. This means that for midpoint and rules, K must always be greater than or equal to the second d the given function, and that for Simpson’s rule, K must always than or equal to the fourth derivative of the given function. In what you’ll be trying to do is find the maximum possible value

Find the error bound ES if n = 4, and then find the number o n that will guarantee the area approximation is accurate withi 1

∫0

2

ex d x

We know that the interval we’re interested in is [a, b] = [0,1] an so plugging these values into the the Simpson’s rule error bo gives ES

K(b − a)5 ≤ 180n 4

ES

K(1 − 0)5 ≤ 180(4)4

ES

K ≤ 46,080

To find a value for K, we’ll need to use the condition that f (4) which means we need to find the fourth derivative of the give 2

f (x) = e x . f (x) = e x

2

2

2

2

f (4)(x) = 12e x + 48x 2e x + 16x 4e x

2

Remember that the interval we’re interested in is [0,1]. Theref to find the maximum value that this fourth derivative can atta interval. Since the fourth derivative is a polynomial function a term is positive, we know it’s increasing throughout the interv means that the largest value it’ll attain is at x = 1. At x = 1, 2

2

f (4)(1) = 12e 1 + 48(1)2e 1 + 16(1)4e 1

2

f (4)(1) = 76e Since this is the largest value the fourth derivative will have d interval, we’ll say K = 76e. ES ≤

76e 46,080

ES ≤ 0.0045 This tells us that the error will be no larger than about 0.0045, Simpson’s rule with n = 4 subintervals to approximate the are curve, we’d get a pretty accurate estimate of actual area. To answer the second part of this question, we need to find t i

i

i

76e(1 − 0)5 ≤ 0.00001 4 180n 76e ≤ 0.0018n 4 76e ≤ n4 0.0018 4

76e ≤n 0.0018

18.41 ≤ n You can’t use 18.41 subintervals, you’d need to use either 18 s or 19 subintervals. But n must be greater than 18.41 in order to an approximation of area within 0.00001 of actual area, which round up and say that n = 19 subintervals. If we were doing this with a midpoint or trapezoidal rule prob stop here. But with Simpson’s rule, remember that we always an even number of subintervals. Which means that we actuall round up to nearest even number, and use n = 20 subintervals guarantee an area estimate within 0.00001....