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and set
z
nC1
D
x: So in order to prove that
z
n
remains negative, we have to show that
(2.39) has a unique negative solution. To this end we define the auxiliary function
f.x/
D
x
tx
2
z
n
;
whereweassumethat
z
n
<0:
First we note that
f
0
.x/
D
1
2tx
and thus f
0
.x/ > 0 for all x<0:Also, f.0/
D
z
n
>0;and we note that f tends
to minus infinity when x tends to minus infinity for any positive value of t: Hence,
as x goes from minus infinity to zero, f is monotonically increasing from minus
infinity to f.0/ > 0; and then it follows that there must be a unique negative value
x
such that
f.x
/
D
0:
This is also illustrated in Fig.
2.4
, where we see that the function x
tx
2
inter-
sects the straight line
z
n
for a uniquely determined negative value of x: It now fol-
lows by induction on n that all values generated by the implicit scheme are negative.
0
−20
−40
−60
−80
−100
−120
−140
−160
−180
−200
−100
−90
−80
−70
−60
−50
−40
−30
−20
−10
0
x
tx
2
Fig. 2.4
The figure depicts the function f.x/
D
x
with t
D
1=100 for the negative
D
values
100
x
0. It illustrates that for any given
horizontal dotted line
f
z
n
,where
z
n
<0,thef
D
f.x/ curve has a single intersection point with the line for x<0
