a + ht+A2h?t2 + A3h³t3 + • .. = f(a)+ f'(a)(t + A2t + A3t° + · .·)+ ½f"(a)(t + A2t (2.138) + Azt +...)² + · .. Equating coefficients of corresponding powers of t gives a = f(a), h = f'(a), Azh? = /½[f"(a) + 2A2f'(a)], Azh3 = 1/6[f"(a) + 6A2f"(a) + 6A3f'(a)], A4h = /24[f" (a)+12A2f'(a)+ (24A3 + 12A3) f"(a) + 24A4 f' (a)], etc. (2.139)
a + ht+A2h?t2 + A3h³t3 + • .. = f(a)+ f'(a)(t + A2t + A3t° + · .·)+ ½f"(a)(t + A2t (2.138) + Azt +...)² + · .. Equating coefficients of corresponding powers of t gives a = f(a), h = f'(a), Azh? = /½[f"(a) + 2A2f'(a)], Azh3 = 1/6[f"(a) + 6A2f"(a) + 6A3f'(a)], A4h = /24[f" (a)+12A2f'(a)+ (24A3 + 12A3) f"(a) + 24A4 f' (a)], etc. (2.139)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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Explain the determaine red
![FIRST-ORDER DIFFERENCE EQUATIONS
67
(i) a = f(a) and 0 < [f'(a)| < 1
From equations (2.125) and (2.127), we have the first-order approximation
Ук+1 — һук —D (1 — Һ)а,
(2.131)
91
where h = f'(a). Therefore,
Yk = a + Ahk.
(2.132)
where A is an arbitrary constant. Let
t = Ahk,
(2.133)
and define
z(t) :
= Yk.
(2.134)
Therefore,
Yk+1 = a + ht,
(2.135)
and from yk+1 = f(yk), we have
a + ht = z(ht) = f[z(t)].
(2.136)
Now assume that z(t) has the asymptotic representation
z(t) = a +t+ A2t + A2h³ + ...
(2.137)
Consequently, we immediately obtain from equation (2.136) the following
result:
a + ht+A2h?t? + A3h³t³ + • ..
= f(a)+ f'(a)(t + A2t + Azt° + .)+/2f"(a)(t + Azt (2.138)
+ A3t + ...
- )² + · ·.
Equating coefficients of corresponding powers of t gives
a = f(a),
h = f'(a),
Azh? = /½[f"(a) + 2A2f'(a)],
Azh3 = 1/6[f"(a) + 6A2ƒ"(a) + 6A3f'(a)],
A4h = 1/24[f"" (a)+ 12A2f'(a) + (24A3 + 12A3) f"(a) + 24A4 f'(a)],
etc.
(2.139)
Solving for A2, A3, A4, etc., gives
f"(a)
A2 :
2(h2 – h)'
|
(h2 – h) f"(a) + 3f"(a)²
A3
6(h³ – h)(h² – h)
|
(3h + 5) f"(a)f"'(a)
12(h4 – h)(h³ – h)
(h + 5) f"(a)³
8(h4 – h)(h³ – h)(h² – h)'
f'' (a)
A4
24(h4 – h)
|
etc.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff127b632-c7db-43de-8096-c44c67b91c3f%2F27d007c4-22b2-4994-a8f1-cb96c214b84c%2F6kzh91b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:FIRST-ORDER DIFFERENCE EQUATIONS
67
(i) a = f(a) and 0 < [f'(a)| < 1
From equations (2.125) and (2.127), we have the first-order approximation
Ук+1 — һук —D (1 — Һ)а,
(2.131)
91
where h = f'(a). Therefore,
Yk = a + Ahk.
(2.132)
where A is an arbitrary constant. Let
t = Ahk,
(2.133)
and define
z(t) :
= Yk.
(2.134)
Therefore,
Yk+1 = a + ht,
(2.135)
and from yk+1 = f(yk), we have
a + ht = z(ht) = f[z(t)].
(2.136)
Now assume that z(t) has the asymptotic representation
z(t) = a +t+ A2t + A2h³ + ...
(2.137)
Consequently, we immediately obtain from equation (2.136) the following
result:
a + ht+A2h?t? + A3h³t³ + • ..
= f(a)+ f'(a)(t + A2t + Azt° + .)+/2f"(a)(t + Azt (2.138)
+ A3t + ...
- )² + · ·.
Equating coefficients of corresponding powers of t gives
a = f(a),
h = f'(a),
Azh? = /½[f"(a) + 2A2f'(a)],
Azh3 = 1/6[f"(a) + 6A2ƒ"(a) + 6A3f'(a)],
A4h = 1/24[f"" (a)+ 12A2f'(a) + (24A3 + 12A3) f"(a) + 24A4 f'(a)],
etc.
(2.139)
Solving for A2, A3, A4, etc., gives
f"(a)
A2 :
2(h2 – h)'
|
(h2 – h) f"(a) + 3f"(a)²
A3
6(h³ – h)(h² – h)
|
(3h + 5) f"(a)f"'(a)
12(h4 – h)(h³ – h)
(h + 5) f"(a)³
8(h4 – h)(h³ – h)(h² – h)'
f'' (a)
A4
24(h4 – h)
|
etc.
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