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Propagation of Errors

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Propagation of Errors

In numerical methods, the calculations are not made with exact numbers. How do these inaccuracies propagate through the calculations?

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Example 1:
Find the bounds for the propagation in adding two numbers. For example if one
is calculating *X *+*Y *where
*X* = 1.5 ± 0.05
*Y* = 3.4 ± 0.04
**Solution
**Maximum possible value of *X* = 1.55 and *Y* = 3.44
Maximum possible value of *X* + *Y* = 1.55 + 3.44 = 4.99
Minimum possible value of *X* = 1.45 and *Y* = 3.36.
Minimum possible value of *X* + *Y* = 1.45 + 3.36 = 4.81
Hence
4.81 ≤ X + Y ≤4.99.

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Propagation of Errors In Formulas

*f nn XXXXX *,,.......,,, 1321 −
*f
*

*n
n
*

*n
n
*

*X
X
fX
*

*X
fX
*

*X
fX
*

*X
ff *∆

∂ ∂

+∆ ∂ ∂

++∆ ∂ ∂

+∆ ∂ ∂

≈∆ − −

1 1

2 2

1 1

.......

If is a function of several variables then the maximum possible value of the error in is

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Example 2:

The strain in an axial member of a square cross- section is given by Given Find the maximum possible error in the measured strain.

*Eh
F
*2∈=

N9.072 ±=*F
*mm1.04 ±=*h
*GPa5.170 ±=*E
*

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Example 2:

)1070()104( 72

923 ×× ∈= −

610286.64 −×= µ286.64=

*E
E
*

*h
h
*

*F
F
*

∆ ∂ ∈∂

+∆ ∂ ∈∂

+∆ ∂ ∈∂

∈=∆

Solution

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Example 2:

*EhF *2
1

= ∂ ∈∂

*Eh
F
*

*h *3
2
−=

∂ ∈∂

22 *Eh
F
*

*E
*−=

∂ ∈∂

*E
Eh
*

*Fh
Eh
FF
*

*Eh
E *∆+∆+∆=∆ 2232

21

9 2923

933923

105.1 )1070()104(

72

0001.0 )1070()104(

7229.0 )1070()104(

1

×× ××

+

× ××

× +×

×× =

−

−−

µ3955.5=

Thus

Hence )3955.5286.64( µµ ±∈=

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Example 3:

Subtraction of numbers that are nearly equal can create unwanted
inaccuracies. Using the formula for error propagation, show that this is true.
**
Solution
**Let
Then
So the relative change is

*yxz *−=

*y
y
zx
*

*x
zz *∆

∂ ∂ +∆

∂ ∂

=∆

*yx *∆−+∆= )1()1(
*yx *∆+∆=

*yx
yx
*

*z
z
*

−

∆+∆ =

∆

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Example 3:

For example if
** **001.02 ±=*x
*

001.0003.2 ±=*y
*

|003.22| 001.0001.0

−

+ =

∆
*z
z
*

= 0.6667 = 66.67%

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