Errors in Numerical Calculation Contents: o o o o o o
Numbers & Their Accuracy Different Types of Errors Mathematical Preliminaries General Error Formula Convergence Assignment 1
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Prepared By Er. Shree Krishna Khadka
NUMBERS & THEIR ACCURACY There are two kinds of numbers: i) Exact numbers, e.g. 1, 2, 3 …1/2, 5/2 etc. ii) Approximate numbers, e.g. PI, K … etc o
Significant Digit: Digits that are used to express a number are called significant digits/figures. The following statements describe the notion of significant digits. - All non-zero digits. - All zeroes occurring between non- zero digits. - Trailing zeroes following a decimal point, e.g. 3.50, 65.0, 0.230 have three
significant digits. - Zeroes between the decimal point are preceding a non-zero digits are not significant, e.g. 0.01234, 0.001234, 0.0001234 all have four significant digits. - When the decimal point is not written, trailing zeroes are not considered to be significant, e.g. 4500 contain only two significant digits. o
Accuracy & precision The concept of accuracy and precision are closely related to significant digits. - Accuracy refers to the number of significant digits in a value. E.g. the number
57.345 is accurate to five significant digits. - Precision refers to the number of decimal position, i.e. the order of
magnitude of last digit in a value. E.g. 57.396 have a precision of 0.001. In numerical computation, we come across numbers which have large numbers of digits and it will be necessary to cut them to usable numbers of figures. This process is known as rounding off. The error caused due to cut-off a large number into usable number of figure is called round-off error. o
Rules for rounding off: To round off a number to n-significant digits, discard all digits to the right of the nth digit if this discarded number is: - Less than half a unit in the n th place, leave the n th digit unaltered. - Greater than half a unit in n th place, increase the n th digit by unity. - Exactly half a unit in the n th place, increase the n th digit by unity if it is odd;
otherwise leave it unchanged. Examples: Following numbers are rounded off to four significant figures. 1.6583(1.658), 30.0567(30.06), 0.859378(0.8594), 3.14159(3.142)
Prepared By Er. Shree Krishna Khadka
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DIFFERENT TYPES OF ERROR:
a) Absolute Error: Numerical difference between true value of a quantity and its approximate value. Mathematically;
−
Absolute Error(EA ) = True Value X
∆
Approximate Value(X1 ) = X
b) Relative error: Ratio of absolute error to a true value of that quantity being concerned. Mathematically;
Relative Error(ER ) =
Absolute Error (EA )
True Value X
c) Percentage Error: The percentage value of relative error. Mathematically;
Percentage Error(EP ) =
Absolute Error (EA )
True Value X
× 100%
d) Truncation Error: Occurs due to truncation or terminating an infinite sequence of operation after a finite number have been performed. Example:
−
cos x = 1
x2 2
,
ex = 1 + x +
x2 2
e) Relative Accuracy: Ratio of change in true value to the modulus of true value. Mathematically;
Relative Accuracy ( R A ) =
Absolute Error (EA )
|True Value X |
~
Absolute Error (EA )
|Approximate Value X 1 |
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Prepared By Er. Shree Krishna Khadka
NUMERICAL EXAMPLES:
An approximate value of PI is given as 3.1428571 and its true value is 3.1415926. Find E A and ER. Solution: EA =| X-X1| = |3.1415926-3.1428571| = 1.265*10 -3 ER = EA/X = 1.265*10 -3/3.1415926 = 4.025*10 -4
Three approximate values of the number 1/3 are given as 0.30, 0.33 and 0.34, which of these is the best approximation? Solution: E1 = |(1/3)-0.30| = 1/30 E2 = |(1/3)-0.33| = 1/300 E3 = |(1/3)-0.34| = 1/150 Since, E2
Evaluate the sum S=_/3 + _/5 + _/7 to four significant digits and find its absolute and relative error. Solution: _/3 = 1.73205 = 1.732 _/5 = 2.23606 = 2.236 _/7 = 2.64574 = 2.646 S = 1.732 + 2.236 + 2.646 = 6.614 Absolute Error (EA) = 0.0005 + 0.0005 + 0.0005 = 0.0015 Relative Error (E R) = EA/S = 0.0015/6.614 = 2.268*10 -4 = 0.0002
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Prepared By Er. Shree Krishna Khadka
MATHEMATICAL PRELIMINARIES
≤ ≤
Theorem 1: If f(x) is continuous in a x b, and if f(a) and f(b) are of opposite signs, then f ω = 0 for at least one number ω, such that a < < .
≤ ≤ ′
Theorem 2 (Rolle’s Theorem): If f(x) is continuous in a x b, f (x) exists in a < < and f(a)=f(b)=0, then there exist at least one value of x say ω, such that f ω = 0.
′
Generalization: If f(x) be a function, which is n-times differentiable on [a, b]. If f(x) vanishes at the (n+1) distinct points x0 , x1 , x2 , , xn in (a,b), then there exist a number ω in (a,b) such that , f n ω = 0.
…
Theorem 3 (Intermediate Value Theorem): If f(x) be continuous in [a, b] and let K be any number between f(a) and f(b), then there exists a number ω in (a, b) such f(ω) = K. Theorem 4 (Mean Value Theorem for Derivative): If f(x) is continuous in [a, b] and f’(x) exists in (a, b), there exists at least one value of x, say ω between a and b such that: f b f(a) f ω = ; a< < b a
−−
′
Theorem 5 (Taylor’s Series for a function of one variable): If f(x) is continuous and passes continuous derivates of order in an interval and includes x=a, then: x a 2 x a n 1 n 1 f x = f a + x a f a + f a + + f a + R n (x) 2! (n 1)!
− − − − ′′ − ⋯ − −
′
Where; R n x : Remainder term =
x a n n!
f n ω ; a <
< .
Theorem 6 (McLaren’s Expansion): 2
x x f x = f 0 + xf ′ 0 + f ′′ 0 + ⋯ + 2!
n
n!
f n 0
Theorem 7 (Taylor’s Series for 2-variables):
∆ ∆ ∂∂ ∆ ∂∂ ∆ ⋯ ∂∂ ∆ ∂ ∆ ∂ ∆ ∂ ∆ ⋯ ⋯ ∂ ∂ ∂
f x1 + x1 , x2 + x2 1 2
{
2 f
x1 2
= f x1 , x2
2
( x1 ) +
2 f
x22
2
+
( x2 ) +
f
x1
+
x1 + 2 f
xn 2
f
x2
x2 +
+
f
xn
xn +
( xn )2 } +
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Prepared By Er. Shree Krishna Khadka
Theorem 8 (Taylor’s Series for several variables):
∆ ∆ ∆ … ∆ … ∂∂ ∆ ∂∂ ∆ ⋯ ∂∂ ∆ ∂ ∆ ∂ ∆ ⋯ ∂ ∆ ∂ ∂ ∂ ∂∂ ∆ ∆ ∂ ∂ ∂ ∆ ∆ … ∂ ∆ − ∆ ⋯ ∂ −∂
f x1 + x1 , x2 + x2 , x3 + x3
xn + xn f f f = f x1 , x2 , x3 xn + x1 + x2 + + x x1 x2 xn n 2 2 2 1 f f f 2 2 + { ( x ) + ( x ) + + ( xn )2 } 1 2 2 2 2 2 x1 x2 xn 2 2 f f +2 x1 x2 + x2 x3 x1 x2 x2 x3 2
+
xn
f
xn
xn
1
1
xn
+
GENERAL ERROR FORMULA Formula for the error committed in using a certain formula or a functional relation. Let; u=f(x1, x2, x3, x4, x5, … xn) be the function of several variable x i (i=1,2…n) and let the error in each xi be xi , then the error u in u is given by:
∆
∆ ∆ ∆ ∆ … ∆
∆
u + u = f ( x1 + x1 , x2 + x2 , x3 + x3
xn + xn ) ….(i)
On expansion of RHS using Taylor’s Series, we get:
∆ ∆ ∆ … ∆ … ∂∂ ∆ ∂∂ ∆ ⋯ ∂∂ ∆ ∂ ∆ ∂ ∆ ⋯ ∂ ∆ ∂ ∂ ∂ ∂∂ ∆ ∆ ∂ ∂ ∂ ∆ ∆ … ∂ ∂− ∂ ∆ − ∆ ⋯ ∆ … ∂∂ ∆ ∆
f x1 + x1 , x2 + x2 , x3 + x3
xn + xn f f f = f x1 , x2 , x3 xn + x1 + x2 + + x x1 x2 xn n 2 2 2 1 f f f 2 2 + { ( x ) + ( x ) + + ( x n )2 } 1 2 2 2 2 2 x1 x2 xn 2 2 2 f f f +2 x1 x2 + x2 x3 + xn 1 xn x1 x2 x2 x3 xn 1 xn +
i.e. u + u = f x1 , x2 , x3
xn
f n i=1 x i
+
xi + term involving ( xi )2
So, ignoring powers of errors, we get:
u=
∂ ∂
∂ ∆x + ∂ ∆x + ⋯ + ∂ ∆x … (ii) ∂ ∂ ∂ ∆ = ∂ ∆ + ∂ ∆ + ⋯ + ∂ ∆ … (iii) = ∂ ∂ ∂
f n i=1 x i
Now, Relative Error: ER
∆
xi = u
u
f
f
1
x1 f
x1
x1 u
f
2
x2
f
x2
x2 u
Prepared By Er. Shree Krishna Khadka
xn
f
xn
xn
u
n
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CONVERGENCE - Numerical computing is based on the idea of iterative process. - Iterative processes involve generation of sequence of approximation with the
hope that, the process will end of the required solution. - Certain methods convert faster than others. - It is necessary to know that convergence rate of any method to get the required solution. - Rapid convergent take less execution time.
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Prepared By Er. Shree Krishna Khadka
ASSIGNMENT 1 Full Marks: 20 Pass Marks: 10 Que: 1 (3X2=6 Marks) 1.1) Round off the following numbers to two decimal places. 2.3742, 81.255, 52.275, 48.21416 1.2) Round off the following numbers to four significant digits. 0.70029, 0.00022218, 2.36425, 38. 46235 1.3)
Calculate (_/102 - _/101) correct to four significant digits.
Que: 2 (2X2=4 Marks) 2.1) if u = 3v 7-6v, find the percentage error in u at v=1, if the error in v is 0.05. 2.2) if y = (0.31x+2.73)/(x+0.35), where the coefficients are rounded off; find the absolute and relative error in y when x= 0.5 ± 0.1.
Que: 3 (3X2=6 Marks) If a = 10.00 ± 0.05, b = 0.0356 ± 0.0002, c = 15300 ± 100 & 500, find the maximum absolute error in: a) a+b+c+d b) a+5c-d c) c3
= ±62000 ±
Que: 4 (1X4=4 Marks) If u=5xy2/z3, x = y = z = 1 & x = y = z = 1;
∆ ∆ ∆
Find the maximum value of relative error. (Hint: Use general error formula)
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Prepared By Er. Shree Krishna Khadka
