Page 264 - Numerical Analysis Using MATLAB and Excel

Taylor and Maclaurin Series

i'' () i''' ()
0
0
2
D
D
3
i ( D v ) D = i () + i' ()v + ---------------v + ----------------v + … (6.135)
0
0
D
D
D
D
D
3!
2!
The first term i () on the right side of (6.135) is found by letting v D = 0 in (6.134). Then,
0
D
i () = I D (6.136)
0
D
To compute the second and third terms of (6.135), we must find the first and second derivatives
of (6.134). These are:
d 1 v D ⁄ nV T 1
i' ( D v ) D = ---------i D = ---------- I e⋅ D and i' () = ---------- I ⋅ D (6.137)
0
D
nV
dv
nV
T
D
T
2
1
1
d
i'' ( D v ) D = -------------i D = ------------- I e⋅ D v D ⁄ nV T and i'' () = ------------- I ⋅ D (6.138)
0
D
2
2
2
d v D n V T 2 n V T 2
Then, by substitution of (6.136), (6.137), and (6.138) into (6.135) we get
⎛ 1 1 ⎞
2
i ( D v ) D = I ⎜ D 1 + ----------v + -------------v + …⎟ (6.139)
2 D
D
2
⎝ nV T n V T ⎠

Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−47
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