Numerical Analysis By Shanker Rao

14 NUMERICAL ANALYSIS
Example 1.13
2
Given that u= 5xy ∆x, ∆y
and ∆z denote the errors in x, y and z respectively such that x = y
3
z
= z = 1 and ∆x = ∆y = ∆z
= 0.001, find the relative maximum error in u.
Solution
We have
2
∂u
5y
∂u
10xy
∂u
xy
= = = − 15
, ,
3 3
4
∂x
z ∂y
z ∂z
z
∂u
∂
∴ ∆u
= ∆x+ u ∂u
∆y
+ ∆z
∂x
∂y
∂z
b g max
∂u
∂
⇒ ∆u
= ∆x+ u ∂u
∆y
+ ∆z
∂x
∂y
∂z
2
2
5y
10xy
= + + − 15xy
∆x
∆y
∆ z
(1)
3 3
4
z z
z
Substituting the given values in (1) and using the formula to find the relative maximum error we get
u
bERg
b∆
gmax 003 .
= = = 0006 . .
max
u 5
Example 1.14
If X = 2.536, find the absolute error and relative error when
(i) X is rounded and
(ii) X is truncated to two decimal digits.
Solution
(i) Here X = 2.536
Rounded-off value of X is x = 2.54
The Absolute Error in X is
E A
= |2.536 – 2.54|
= |– 0.004| = 0.004
Relative Error = E R
= 0 . 004 = 0.0015772
2.
536
= 1.5772 × 10 –3 .
(ii) Truncated Value of X is x = 2.53
Absolute Error E A
= |2.536 – 2.53| = |0.006| = 0.006
2
∴
Relative Error = E R
= E A
X
= 0 . 006
2.
536
= 0.0023659
= 2.3659 × 10 –3 .
Example 1.15 If π= 22
7
22
Solution Absolute Error = E A
= −
7
is approximated as 3.14, find the absolute error, relative error and relative percentage error.
314 . =
22 − 2198 .
7