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Numerical Methods:
Introduction, Accuracy, Errors
ADVANCED ENGINEERING MATHEMATICS: LESSON 5
PREPARED BY: ENGR. APRIL JOY F. AGUADO
COLLEGE OF ENGINEERING AND ARCHITECTURE
What is a Numerical Analysis?
➒ Numerical Analysis is a branch of
mathematics which deals with the
approximate solutions of
mathematical problems.
2
What are numerical methods?
➒ Are methods of obtaining solution by subjecting the
original problem to a series of steps or repetitions.
➒ Numerical Methods develop accurate and fast
approximations to problems whose exact solutions
are difficult to find because of their complexity.
3
Common Ways to Express Error
1. Absolute Error (AE) = π‘₯ βˆ’ π‘₯β€²
where π‘₯β€² = is the approximate value
π‘₯ = is the exact value
2. Relative Error (RE) =
π‘₯βˆ’π‘₯β€²
π‘₯
=
𝐴𝐸
π‘‡π‘Ÿπ‘’π‘’ π‘‰π‘Žπ‘™π‘’π‘’
3. Percentage Error = RE x 100
4
Accuracy vs Precision
➒ Accuracy – is the closeness of values to its true
value.
➒ Precision – is the closeness of values to each
other.
5
Sources of Error
➒ Modeling Error
o Blunders
o Formulation Error
o Data Uncertainty
➒ Numerical Error
o Round-off Error
o Truncation Error
6
Round-off Error
➒ Round-off error – error created due to
approximate representation of numbers
7
Truncation Error
➒ Truncation Error – is an error due to truncating a process involving
infinite number of steps to finite number of steps
➒ Error due to numerical approximation method
o Limiting infinite series to finite number of terms
o Limiting infinite number of iterations for finite number of
iterations
o Taking finite step size instead of infinitesimal step size (numerical
differentiation and numerical integration)
8
Truncating an infinite series
➒ Maclaurin series of 𝑒π‘₯ = 1 +
π‘₯
1
+
π‘₯2
2!
+
π‘₯3
3!
+
π‘₯4
4!
+ β‹―
➒ Use the terms on the right side to determine the value of 𝑒π‘₯
➒ We cannot use infinite terms: say we use only first four terms
➒ Approximate 𝑒π‘₯
by 𝑒π‘₯
β‰ˆ 1 +
π‘₯
1
+
π‘₯2
2!
+
π‘₯3
3!
➒ Truncation Error = 𝑒π‘₯
βˆ’
π‘₯
1
+
π‘₯2
2!
+
π‘₯3
3!
=
π‘₯4
4!
+
π‘₯5
5!
+ β‹―
9
Truncating an infinite series
➒ Illustration: Estimate 𝑒0.5
for different number of terms and calculate
the absolute error (exact value of 𝑒0.5 up to 5 decimal places is
1.64872).
10
Number of
Terms
Estimate of
𝑒0.5
Absolute
Error
1 1 0.64872
2 1.5 0.14872
3 1.625 0.012372
4 1.645832 0.00289
5 1.64843617 0.00028
Truncation Error Integration
11
Truncation Error in Integration
➒ ‫׬‬2
4
π‘₯2
𝑑π‘₯ =
π‘₯3
3
in 2 to 4 =
43βˆ’23
3
=
64βˆ’8
3
= 18.67 gives exact value
rounded to 2 decimal places
➒ Let us apply numerical approximation method:
o For simplicity, ease and convenience and to avoid round-off errors,
take rectangles of width 1 from 2 to 4 and sum up their areas as an
estimate of the integral ‫׬‬2
4
π‘₯2 𝑑π‘₯ giving us
o Estimate value 𝐼a = 1 4 + 1 9 = 13 (Height 22 and 42)
o Truncation Error = 18.67-13=5.67
12
Truncation Error Integration
13
Truncation Error in Integration
➒ What would happen, if we rework the same example with smaller
width? (smaller size and increase the number of steps)
➒ Let number of subintervals now in [2,4] be 4 instead of just 2
➒ Width of each rectangle would be now 0.5
➒ Heights would be 4, 6.25, 9, 12.5
➒ πΌπ‘Ž = 0.5 4 + 6.25 + 9 + 12.25 = 15.75
➒ Truncation error = 18.67-15.75=2.92 (reduction from 5.67 to 2.92)
14
Truncation Error Integration
15
Truncation Error in Differentiation
➒ Take 𝑓 π‘₯ = π‘₯3
and let us estimate derivative at π‘₯ = 2
➒ 𝑓′
π‘₯ = 3π‘₯2
; 𝑓′
2 = 3 22
= 12 (true value)
➒ The derivative is defines as 𝑓′
π‘₯ = lim
β„Žβ†’0
𝑓 π‘₯+β„Ž βˆ’π‘“(π‘₯)
β„Ž
➒ Let us estimate derivative of 2 by taking β„Ž = 0.1
➒ 𝑓′
2 β‰…
𝑓 2.1 βˆ’π‘“ 2
0.1
=
2.13βˆ’23
0.1
= 12.61 giving an absolute error of 0.61
16
Truncation Error Differentiation
17
Truncation Error in Differentiation
➒ If we reduce β„Ž to 0.05 and repeat the same exercise, estimate
obtained is
➒ 𝑓′
2 β‰…
𝑓 2.05 βˆ’π‘“ 2
0.05
=
2.053βˆ’23
0.1
= 12.3025 giving an absolute error of
0.3025
18
Observation
➒ Truncation error
o Arises due to numerical approximation method being applied to
solve the problem
o Arises basically due to truncating the process to finite number of
steps
o As step size is reduce, truncation error decreases
o Reduction in step size ↔ increase in number of steps
19
20
ALL is WELL. TRUST the PROCESS.
KEEP FIGHTING, FUTURE ENGINEERS!
-
Ma’am A

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Numerical Methods: Accuracy, Errors & Truncation

  • 1. Numerical Methods: Introduction, Accuracy, Errors ADVANCED ENGINEERING MATHEMATICS: LESSON 5 PREPARED BY: ENGR. APRIL JOY F. AGUADO COLLEGE OF ENGINEERING AND ARCHITECTURE
  • 2. What is a Numerical Analysis? ➒ Numerical Analysis is a branch of mathematics which deals with the approximate solutions of mathematical problems. 2
  • 3. What are numerical methods? ➒ Are methods of obtaining solution by subjecting the original problem to a series of steps or repetitions. ➒ Numerical Methods develop accurate and fast approximations to problems whose exact solutions are difficult to find because of their complexity. 3
  • 4. Common Ways to Express Error 1. Absolute Error (AE) = π‘₯ βˆ’ π‘₯β€² where π‘₯β€² = is the approximate value π‘₯ = is the exact value 2. Relative Error (RE) = π‘₯βˆ’π‘₯β€² π‘₯ = 𝐴𝐸 π‘‡π‘Ÿπ‘’π‘’ π‘‰π‘Žπ‘™π‘’π‘’ 3. Percentage Error = RE x 100 4
  • 5. Accuracy vs Precision ➒ Accuracy – is the closeness of values to its true value. ➒ Precision – is the closeness of values to each other. 5
  • 6. Sources of Error ➒ Modeling Error o Blunders o Formulation Error o Data Uncertainty ➒ Numerical Error o Round-off Error o Truncation Error 6
  • 7. Round-off Error ➒ Round-off error – error created due to approximate representation of numbers 7
  • 8. Truncation Error ➒ Truncation Error – is an error due to truncating a process involving infinite number of steps to finite number of steps ➒ Error due to numerical approximation method o Limiting infinite series to finite number of terms o Limiting infinite number of iterations for finite number of iterations o Taking finite step size instead of infinitesimal step size (numerical differentiation and numerical integration) 8
  • 9. Truncating an infinite series ➒ Maclaurin series of 𝑒π‘₯ = 1 + π‘₯ 1 + π‘₯2 2! + π‘₯3 3! + π‘₯4 4! + β‹― ➒ Use the terms on the right side to determine the value of 𝑒π‘₯ ➒ We cannot use infinite terms: say we use only first four terms ➒ Approximate 𝑒π‘₯ by 𝑒π‘₯ β‰ˆ 1 + π‘₯ 1 + π‘₯2 2! + π‘₯3 3! ➒ Truncation Error = 𝑒π‘₯ βˆ’ π‘₯ 1 + π‘₯2 2! + π‘₯3 3! = π‘₯4 4! + π‘₯5 5! + β‹― 9
  • 10. Truncating an infinite series ➒ Illustration: Estimate 𝑒0.5 for different number of terms and calculate the absolute error (exact value of 𝑒0.5 up to 5 decimal places is 1.64872). 10 Number of Terms Estimate of 𝑒0.5 Absolute Error 1 1 0.64872 2 1.5 0.14872 3 1.625 0.012372 4 1.645832 0.00289 5 1.64843617 0.00028
  • 12. Truncation Error in Integration ➒ ‫׬‬2 4 π‘₯2 𝑑π‘₯ = π‘₯3 3 in 2 to 4 = 43βˆ’23 3 = 64βˆ’8 3 = 18.67 gives exact value rounded to 2 decimal places ➒ Let us apply numerical approximation method: o For simplicity, ease and convenience and to avoid round-off errors, take rectangles of width 1 from 2 to 4 and sum up their areas as an estimate of the integral ‫׬‬2 4 π‘₯2 𝑑π‘₯ giving us o Estimate value 𝐼a = 1 4 + 1 9 = 13 (Height 22 and 42) o Truncation Error = 18.67-13=5.67 12
  • 14. Truncation Error in Integration ➒ What would happen, if we rework the same example with smaller width? (smaller size and increase the number of steps) ➒ Let number of subintervals now in [2,4] be 4 instead of just 2 ➒ Width of each rectangle would be now 0.5 ➒ Heights would be 4, 6.25, 9, 12.5 ➒ πΌπ‘Ž = 0.5 4 + 6.25 + 9 + 12.25 = 15.75 ➒ Truncation error = 18.67-15.75=2.92 (reduction from 5.67 to 2.92) 14
  • 16. Truncation Error in Differentiation ➒ Take 𝑓 π‘₯ = π‘₯3 and let us estimate derivative at π‘₯ = 2 ➒ 𝑓′ π‘₯ = 3π‘₯2 ; 𝑓′ 2 = 3 22 = 12 (true value) ➒ The derivative is defines as 𝑓′ π‘₯ = lim β„Žβ†’0 𝑓 π‘₯+β„Ž βˆ’π‘“(π‘₯) β„Ž ➒ Let us estimate derivative of 2 by taking β„Ž = 0.1 ➒ 𝑓′ 2 β‰… 𝑓 2.1 βˆ’π‘“ 2 0.1 = 2.13βˆ’23 0.1 = 12.61 giving an absolute error of 0.61 16
  • 18. Truncation Error in Differentiation ➒ If we reduce β„Ž to 0.05 and repeat the same exercise, estimate obtained is ➒ 𝑓′ 2 β‰… 𝑓 2.05 βˆ’π‘“ 2 0.05 = 2.053βˆ’23 0.1 = 12.3025 giving an absolute error of 0.3025 18
  • 19. Observation ➒ Truncation error o Arises due to numerical approximation method being applied to solve the problem o Arises basically due to truncating the process to finite number of steps o As step size is reduce, truncation error decreases o Reduction in step size ↔ increase in number of steps 19
  • 20. 20 ALL is WELL. TRUST the PROCESS. KEEP FIGHTING, FUTURE ENGINEERS! - Ma’am A