g (x ) ≈ g (0 ) + g ′(0 )(x − a ) = 0 + φx = φx .
Eq. 33
It is important to note that since φ gives the failure rate, the approximation assumes that the failure rate is linear in x (see Fig. 4: Left Panel). Using Equation 33 in the numerator in Equation 31, we can write the following expression, which employs another first-order Taylor approximation justified below: m s (k, φ, x ) ≈ φλjce e −λ
− φx
k (λφx ) i−1 (λφx ) i−1 −λ = φλjce ∑ . ∑ − φx i =1 (e i =1 (i − 1) ! )(i − 1)! k
Eq. 34
On comparing Equations 31 and 34, it would seem that the critical part of this last
0.2
1.2
0.18
1.15
0.16
1.1 0.14
1.05 f(x)
f(x)
0.12 0.1 0.08
1 0.95
0.06
0.9 0.04
0.85
0.02
0.8
0 20
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50
60
70
20
80
30
40
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Age (years)
Age (years)
Figure 4. First-order Taylor approximations. Left Panel: the function g (x ) = φx (solid
line) is a good approximation of the function g(x ) = 1 − e
(
φx << 1 , the function g(x ) = ⎡1 − 1 − e ⎢⎣ − φx (diamonds). function g(x ) = e
−φ x
− φx
(diamonds). Right Panel: for
) ⎤⎥⎦ (solid line) is well approximated by the i
18