A textbook of Computer Based Numerical and Statiscal Techniques part 3 ppsx

6
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Using the principle of equal effects, which states
12
1
2
X
X
xx
x
x
∂∂
δ=δ
∂∂
= =
n
n
X
x
x
∂
δ
∂
this
implies that
1
1
∂
δ=δ
∂
X

Xnx
x
or
1
1
X
x
X
n
x
δ
δ=
∂
∂
. Similarly we get
23
23
,
,
XX
xx
XX
nn
xx
δδ
δ= δ=
∂∂
∂∂
,
n

n
X
x
X
n
x
δ
δ=
∂
∂
and so on.
This form is useful where error in dependent variable is given and also we are to find errors
in both independent variables.
Remark: The Error
1
10
2
n
−
=×
, if a number is correct to n decimal places. Also Relative
error is less than
1
1
10
n
l
−
×
, if number is correct to n significant digits and l is the first significant

digit of a number.
1.4.6 Error in Evaluating
x
k
Let x
k
be the function having k is an integer or fraction then Relative Error for this function is
given
Relative Error =
or
xXx
kk
xXx
δδ δ
≤
Example 2. Find the absolute, percentage and relative errors if x is rounded-off to three decimal
digits. Given x = 0.005998.
Sol. If x is rounded-off to three decimal places we get x = 0.006. Therefore
Error = True value – Approximate value
Error = .005998 – .006 = – .000002
Absolute Error = E
a
=
Error
= 0.000002
Relative Error =
0.000002
0.0033344
0.005998 0.005998
aa

r
EE
E
True value
= ===
and
Percentage Error = E
p
= E
r
× 100 = 0.33344.
Example 3. Find the number of trustworthy figure in (0.491)
3
assuming that the number 0.491 is
correct to last figure.
Sol. We know that Relative Error, E
r
=
δδ
≤
Xx
k
Xx
Here δx = 0.0005 because
−
×=
3
1
10 0.0005
2

ERRORS AND FLOATING POINT
7
or
3
0.0005 3 0.0005
3 0.01267
0.118371
(0.491)
x
k
x
×
=× = =
δ
Therefore, Absolute Error = E
r
.X
or Absolute Error < 0.01267 × (0.491)
3
= 0.01267 × 0.118371
= 0.0015
The error affects the third decimal place, therefore, (0.491)
3
= 0.1183 is correct to second
decimal places.
Example 4. If 0.333 is the approximate value of
1
3
, then find its absolute, relative and percentage
errors.

Sol. Given that True value
()
1
3
x
=
, and its Approximate value (x′) = 0.333
Therefore, Absolute Error,
1
0.333 0.333333 0.333 0.000333
3
a
Exx
′
=− =− = − =
Relative Error, E
r
=
0.000333
0.000999
0.333333
a
E
x
==
and
Percentage Error, E
p
=
100 0.000999 100 0.099%

r
E
×= ×=
.
Example 5. Round-off the number 75462 to four significant digits and then calculate its absolute
error, relative error and percentage error.
Sol. After rounded-off the number to four significant digits we get 75460.
Therefore Absolute Error E
a
=
75462 75460 2−=
Relative Error E
r
=
2
75462 75462
aa
EE
true value
==
= 0.0000265
Percentage Error E
p
= E
r
× 100 = 0.00265.
Example 6. Find the relative error of the number 8.6 if both of its digits are correct.
Sol. Since
−
=× =

1
1
10 0.05
2
a
E
therefore, Relative Error =
==
0.5
0.0058
8.6
r
E
.
Example 7. Three approximate values of number
1
3
are given as 0.30, 0.33 and 0.34. Which of
these three is the best approximation?
Sol. The number, which has least absolute error, gives the best approximation.
True value x =
1
0.33333
3
=
When approximate value x′ is 0.30 the Absolute Error is given by:
E
a
= −= − =
′

0.33333 0.30 0.03333xx
8
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
When approximate value x′ is 0.33 the Absolute Error is given by:
E
a
=
′
−= − =0.33333 0.33 0.00333xx
When approximate value x′ is 0.34 the Absolute Error is given by:
E
a
=
′
−= − =0.33333 0.34 0.00667xx
Here absolute error is least when approximate value is 0.33. Hence 0.33 is the best
approximation.
Example 8. Calculate the sum of
3, 5
and
7
to four significant digits and find its absolute
and relative errors.
Sol. Here
===
3 1.732, 5 2.236, 7 2.646
Hence Sum = 6.614 and
Absolute Error = E
a
= 0.0005 + 0.0005 + 0.0005 = 0.0015

(Because
−
×
3
1
10 = 0.0005).
2
Also the total absolute error shows that the sum is correct up
to 3 significant figures. Therefore S = 6.61 and
Relative Error,E
r
=
=
0.0015
0.0002
6.61
.
Example 9. Approximate values of
1
7
and
1
11
, correct to 4 decimal places are 0.1429 and 0.0909
respectively. Find the possible relative error and absolute error in the sum of 0.1429 and 0.0909.
Sol. The maximum error in each case =
−
×=× =
4
11

10 0.0001 0.00005
22
1. Relative Error, E
r
=
δ
<+
0.00005 0.00005
0.2338 0.2338
X
X
(Because X = 0.1429 + 0.0909)
Therefore,
δ
<=
0.0001
0.00043
0.2338
X
X
2. Absolute Error, E
a
=
0.0001
0.2338 0.0001.
0.2338
X
X
X
δ

×= × =
Example 10. Find the number of trustworthy figures in (367)
1/5
where 367 is correct to three
significant figures.
Sol. Relative Error
δ
<
1
;
5
r
x
E
x
Therefore,
δ
=× =
1 1 0.5
0.0003
5 5 367
x
x
Similarly, Absolute Error
<×=×=
1/5
(367) 0.0003 3.258 0.0003 0.001
a
E
Hence Absolute Error <0.001.

Thus error effects fourth significant figure and hence (367)
1/5
≈ 3.26 correct to the three
figures.
ERRORS AND FLOATING POINT
9
Example 11. Find the relative error in calculation of
7.342
0.241
. Where numbers 7.342 and 0.241 are
correct to three decimal places. Determine the smallest interval in which true result lies.
Sol. Relative Error
δδ
≤+
12
12
xx
xx
Here δx
1
= δx
2
= 0.0005, x
1
= 7.342, x
2
= 0.241
Therefore, Relative Error
≤+
0.0005 0.0005

7.342 0.241
×

≤+= =

×

1 1 0.0005 7.583
0.0005 0.0021
7.342 0.241 7.342 0.241
Similarly, Absolute Error

×
≤×= =


1
2
0.0021 7.342
0.0021 0.0639
0.241
x
x
Here
1
2
x
x
=
=

7.342
30.4647
0.241
Hence true value of
7.342
0.241
lies between 30.4647 – 0.0639 = 30. 4008 and 30.4647 + 0.0639 =
30.5286.
Example 12. Find the product of 346.1 and 865.2 and state how many figures of the result are
trustworthy, given that the numbers are correct to four significant figures.
Sol. For given numbers 346.1 and 865.2,
δx
1
= 0.05 = δx
2
Because Error =
−
×
1
10
2
n
Also, X = 346.1 × 865.2 = 299446 (correct to six significant figures)
Therefore Relative Error
δδ
≤+= +
12
12
0.05 0.05
346.1 865.2

r
xx
E
xx
= 0.000144 + 0.000058 = 0.000202
Similarly, Absolute Error E
a

= E
r
X ≤ 0.000202 × 299446
≈
60
So, true value of the product of the given numbers lies between
299446 – 60 = 299386 And 299446 + 60 = 299506.
Hence the mean of these values is
+
=
299386 299506
299446
2
which is written as 299.4 × 10
3
.
This is correct to four significant figures.
Example 13. Find the relative error in the calculation of 3.724 × 4.312 and determine the interval
in which true result lies. Given that the numbers 3.724 and 4.312 are correct to last digit?
Sol. For product of numbers, Relative Error =
δδδ
δ

≤++ +
12
12

n
n
xxx
X
Xx x x
10
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Therefore, Relative Error, E
r
=

+=
0.0005 0.0005
0.0002501
3.724 4.312
(Because Error
−−
=× =× =
3
11
10 10 0.0005)
22
n
Absolute Error, E
a
= E

r
X = 0.002501 × 3.724 × 4.312
= 0.0040157
Product x
1
x
2
= 3.724 × 4.312 = 16.057888
Lower limit is given by 16.057888 – 0.004016 = 16.053872
Upper limit is given by 16.057888 + 0.004016=16.061904
Hence true value lies between 16.0539 and 16.0619.
Example 14. Find the absolute error in calculating (768)
1/5
and determine the interval in which true
value lies 768 is correct its last digit.
Sol. Relative Error, E
r
=
δx
k
x
=×
10.5
5 768
=
=
0.1
0.0001302
768
Absolute Error, E

a
= E
r
× (768)
1/5
= 0.0001302 × 3.77636
= 0.0004916
Therefore, lower limit = 3.77636 – 0.00049 = 3.77587 and
Upper limit = 3.77636 + 0.00049 = 3.77685
Hence value of (768)
1/5
lies between 3.77587 and 3.77685.
Example 15. Find the number of correct figure in the quotient
65.3
7
, assuming that the numerator
is correct to last figure.
Sol. Since Relative Error,
12
12
δδ
≤+
r
xx
E
xx
Here δx
1
= 0.05 = δx
2

, x
1
= 65.7 and x
2
= 2.6
Therefore, Relative Error,
0.05 0.05 0.05 68.3
0.01999
65.7 2.6 65.7 2.6
r
E
×
≤+≤ =
×
Also, Absolute Error,
65.3
0.01999 0.502
2.6
a
E
≤×=
(since the error affects the first decimal
place).
Example 16. Find the percentage error if 625.483 is approximated to three significant figures.
Sol. Here x = 625.483 and x′ = 625.0 therefore,
Absolute Error,
625.483 625 0.483,
a
E
=−=

ERRORS AND FLOATING POINT
11
Relative Error,
0.483
0.000772
625.483 625.483
a
r
E
E
===
and
Percentage Error,
100 0.077%.
pr
EE
=× =
Example 17. Find the relative error in taking the difference of numbers
5.5 = 2.345
and
6.1 = 2.470
. Numbers should be correct to four significant figures.
Sol. Relative Error
12
r
xx
E
XX
δδ
≤+

Here δx
1
= 0.0005 = δx
2
Therefore, Relative Error =
1
0.0005
22
2.470 2.345
x
X
δ
=
−
=
0.0005 0.001
20.008
0.125 0.125
==
.
Example 18. If X = x + e prove that
−≈
e
Xx
2X
.
Sol. L.H.S.
Xx−
=
1

2
1
e
XXe XX
x

−−= − −


=
1
2
e
XX
X

−−


=
22
ee
XX
XX
−+ ≈
. R.H.S. proved.
Example 19. If
23
4
4x y

u=
z
and errors in x, y, z be 0.001, compute the relative maximum error in
u when x = y = z = 1.
Sol. We know
uuu
uxyz
xyz
∂∂∂
δ= δ+ δ+ δ
∂∂∂
Since
−
∂∂ ∂
== =
∂∂ ∂
322 23
44 5
812 16
,,
xy xy xy
uu u
xy z
zz z
Also the errors δx, δy, δz may be positive or negative, therefore absolute values of terms on
R.H.S. is,
()
δ
max
u

=
δ+ δ+ δ
32223
44 5
812 12
xy xy xy
xyz
zz z
= 8(0.001) + 12(0.001) + 16(0.001) = 0.036
Also, Max. Relative Error =
=
0.036
0.009
4
(Because E
r(max)
=
δ
;
u
u
u = 4 at x = y = z = 1).
12
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Example 20. It is required to obtain the roots of X
2
– 2X + log
10
2 = 0 to four decimal places. To
what accuracy should log

10
2 be given?
Sol. Roots of the equation X
2
– 2X + log
10
2 = 0 are given by,
X =
±−
=± −
10
10
244log2
11log2
2
Therefore, δX =
4
(log 2)
1
0.5 10
2
1log2
−
δ
<×
−
or
−−
δ<××− < ×
41/2 4

(log 2) 2 0.5 10 (1 log 2) 0.83604 10
≈
8.3604 × 10
–5
.
Example 21. If r = 3h(h
6
– 2), find the percentage error in r at h = 1, if the percentage error in
h is 5.
Sol. We know δ
n
x
=
δ
∂
∂
n
X
X
n
x
where X = f(x
1
, x
2
, , x
n
)
Therefore,
δr

=
δ
δ= − δ
∂
6
(21 6)
r
hh h
h
δ
×100
r
r
=

−
δ×

−

6
7
21 6
100
36
h
h
hh
=
−δ



×=×=−


−−
 
21 6 15
100 5 25%
36 3
h
h
Percentage Error =
δ
=× =
100 25%
p
r
E
r
.
Example 22. Find the relative error in the function
12
12

n
mmm
n
yax x x
=

.
Sol. Given function
12
12

n
mmm
n
yax x x
=
.
On taking log both the sides, we get
log y = log a + m
1
log x
1
+ m
2
log x
2
+ + m
n
log x
n
Therefore,
 
∂∂∂
===
 
∂∂∂

  
312
11 22 33
11 1
, , etc.
yyym
mm
yx xyx xyx x
Hence Relative Error,
∂∂ ∂
δδδ
=++ +
∂∂ ∂
12
12

n
r
n
yy y
xxx
E
xy x y x y
=
δδδ
+++
12
12
12

n
n
n
xxx
mm m
xx x
ERRORS AND FLOATING POINT
13
Since errors δx
1
, δx
2
may be positive or negative, therefore absolute values of terms on
R.H.S. give,
()
δ
δδ
≤+++
12
12
max
12

n
rn
n
x
xx
Emm m
xx x

.
Remark: If y = x
1
x
2
x
3
x
n
, then relative error is given by
δδδ
≈++ +
12
12

n
r
n
xxx
E
xx x
Therefore relative error of n product of n numbers is approximately equal to the algebraic
sum of their relative errors.
Example 23. The discharge Q over a notch for head H is calculated by the formula
Q = kH
5/2
, where k is a given constant. If the head is 75 cm and an error of 0.15 cm is possible in its
measurement, estimate the percentage error in computing the discharge.
Sol. Given that Q = kH
5/2

On taking Log both the sides of the equation, we have
log Q = log
+
5
2
k
log H
On differentiating, we get
5
2
QH
QH
δδ
=
50.15 1
100 100 0.5.
275 2
Q
Q
δ
×=× ×==
Example 24. Compute the percentage error in the time period
=
1
T2
g
π
for l = 1m if the error
in the measurement of l is 0.01.
Sol. Given T =

π2
l
g
On taking log both the sides, we get
log T = log 2π +
−
11
log log
22
lg
⇒
δ
1
T
T
=
δ1
2
l
l
⇒
δ
×100
T
T
=
δ
×= ×=
×
0.01

100 100 0.5%
221
l
l
.
Example 25. If u = 2V
6
– 5V, find the percentage error in u at V = 1, if error in V is 0.05.
Sol. Given u = 2V
6
– 5V
δu =
∂
δ= −δ
∂
5
(12 5)
u
VV V
V
14
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
∂
×100
u
u
=

−
δ×


−
5
6
12 5
100
25
V
V
VV
=
−
××=−×=−
−
(12 5) 7
0.05 100 5 11.667%
(2 5) 3
Hence maximum percentage error (E
p
)
max
= 11.667%.
Example 26. How accurately should the length and time of vibration of a pendulum should be
measured in order that the computed value of g is correct to 0.01%.
Sol. Period of vibration T is given by T =
π2
l
g
, where l is the length of pendulum.
Therefore, g =

∂
ππ
⇒=
∂
22
22
44
g
l
l
TT
and
()
∂
π
=−
∂
2
3
4
2
g
l
T
T
δl
=
and
22
gg

T
gg
lT
δδ
δ=
∂∂
 
 
∂∂
 
(1)
But the percentage error in g is
δ
×100
g
g
= 0.01 ⇒
2
2
100 0.01
4
g
l
T
δ
×=
π
(2)
(a) Percentage Error in l =
δ

×100
l
l
=
1
100 Because
2
2
gg
l
g
g
l
l
l
δδ
  
×δ=


∂
∂




∂

∂




=
() ()
22 2 2
11
100 100
2
24 4
gg
l
TlT

δδ

×= ×

ππ

=
×=
1
0.01 0.005
2
(From 2)
(b) Percentage Error in T=
100
T
T
×

δ
=
1
100
2
g
TgT
δ

×

∂∂

=
2
3
100
4
4
g
l
T
δ
×

π


=
×=

1
100 0.0025
4
. (From 2)
ERRORS AND FLOATING POINT
15
Example 27. Calculate the value of x – x cos θ correct to three significant figures if x = 10.2 cm,
and θ = 5°. Find permissible errors also in x and θ.
Sol. Given that θ = 5° =
π
=
511
radian
180 126
1 – cos θ =

θθ
−− + −


24
1 1
2! 4!
=
θθ
 
−+ = − +
 
 
24

24
111 1 11

2! 4! 2 126 24 126
= 0.0038107 – 0.0000024
≈
0.0038083
Therefore X = x(1 – cos θ) = 10.2(0.0038083) = 0.0388446 ≈ 0.0388
Further, δx =
δ
=≈
∂
×



∂
0.0005
0.0656
2 0.0038083
2
X
X
x
δθ =
δ
==
∂
θ× ×




∂θ
0.0005 0.0005
2 sin 2 10.2 0.0871907
2
X
X
x
where sin θ =
θθ

θ−+− = − + =


3
35
11 1 11
0.0871907
3! 5! 126 6 126
Therefore δθ = ≈≈
×
0.0005
0.0002809 0.00028
20.4 0.0871907
.
Example 28. The percentage error in R which is given by
=+
2
rh

R
2h 2
, is not allowed to exceed
0.2%, find allowable error in r and h when r = 4.5 cm and h = 5.5 cm.
Sol. The percentage error in R =
δ
×=100 0.2
R
R
Therefore δR =
()
2
4.5
0.2 0.2 5.5
100 100 2 5.5 2
R


×= +

×

Because R =
+
2
22
rh
h
=
×

×=
0.2 50.50 0.002 50.50
100 11 11
(1)