(7) k +1\-(k+1) Ck. Ck+1 = e (7.93) Note that for k → 0, this equation becomes Ck+1 ~ Ck, (7.94) which implies that Ch ~ constant. Hence, in order to determine the dominant behavior of c, higher-order terms must be included in the k dependent factor on the right side of equation (7.93). To proceed, define A(k) as follows: -(k+1) Aa) - (1+) = e (7.95) Therefore, for k o, we have log A = log(e) – (k +1) log ( 1+ %3D k 1 1 1 – (k +1) ( (7.96) = 1 - k 2k2 3k3 4k4 1 1 1 2k 6k2 12k3 and 1 1 1 A(k) = exp 2k 6k2 12k3 (7.97) 1 1 2k 6k2 12k3 If only terms to order k-3 are retained, then 1 1 1 1 1 1 1 A(k) = 1+ +.. 2k 6k2 12k3 4k2 6k3 8k3 1 3 +.. 1- + 2k 24k2 16k3 (7.98) Hence, Cr obeys the difference equation (1 1 7 3 +... (7.99) Ck+1 = Ck. 2k 24k2 16k3

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine yellow

226
Difference Equations
then
()
k
k
ak
(7.92)
e
Again, substitution of equation (7.92) into equation (7.80) gives
- (k+1)
(*)
k +1
Ck+1
Ck.
(7.93)
= e
k
Note that for k → 0, this equation becomes
Ck+1 ~ Ck,
(7.94)
which implies that c ~ constant. Hence, in order to determine the dominant
Ck, higher-order terms must be included in the k dependent factor
behavior of
on the right side of equation (7.93).
To proceed, define A(k) as follows:
-(k+1)
A(k) = e
1+
(7.95)
Therefore, for k → 0, we have
log A = log(e) – (k +1) log ( 1+
1
= 1- (k + 1)
k
1
1
1
(7.96)
2k2
3k3
4k4
1
1
1
2k
6k2
12k3
and
1
1
1
A(k)
exp
2k
6k2
12k3
(7.97)
1
1
1
2k
6k2
12k3
r=
If only terms to order k-3 are retained, then
1
+
2k
1
1
1
1
1
A(k)
1+
+..
6k2
12k3
2
4k2
6k3
8k3
1
7
1
2k
3
+...
24k2
16k3
(7.98)
Hence, Ck obeys the difference equation
:).
1
7
1
2k
(7.99)
Ck+1 =
+...
Ck•
24k2
16k3
APPLICATIONS
227
Transcribed Image Text:226 Difference Equations then () k k ak (7.92) e Again, substitution of equation (7.92) into equation (7.80) gives - (k+1) (*) k +1 Ck+1 Ck. (7.93) = e k Note that for k → 0, this equation becomes Ck+1 ~ Ck, (7.94) which implies that c ~ constant. Hence, in order to determine the dominant Ck, higher-order terms must be included in the k dependent factor behavior of on the right side of equation (7.93). To proceed, define A(k) as follows: -(k+1) A(k) = e 1+ (7.95) Therefore, for k → 0, we have log A = log(e) – (k +1) log ( 1+ 1 = 1- (k + 1) k 1 1 1 (7.96) 2k2 3k3 4k4 1 1 1 2k 6k2 12k3 and 1 1 1 A(k) exp 2k 6k2 12k3 (7.97) 1 1 1 2k 6k2 12k3 r= If only terms to order k-3 are retained, then 1 + 2k 1 1 1 1 1 A(k) 1+ +.. 6k2 12k3 2 4k2 6k3 8k3 1 7 1 2k 3 +... 24k2 16k3 (7.98) Hence, Ck obeys the difference equation :). 1 7 1 2k (7.99) Ck+1 = +... Ck• 24k2 16k3 APPLICATIONS 227
l stc ksa
2:37 PM
C @ 1 65% 4
equation
ak+1 =
= kak.
(7.80)
This follows directly from the results of Section 2.4. To determine the dom-
inant or leading behavior of the solution to equation (7.80), we rewrite it in
the form
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Transcribed Image Text:l stc ksa 2:37 PM C @ 1 65% 4 equation ak+1 = = kak. (7.80) This follows directly from the results of Section 2.4. To determine the dom- inant or leading behavior of the solution to equation (7.80), we rewrite it in the form Cancel Actual Size (410 KB) Choose
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