Yk+1 = f(a) + f^(a)(uk)", (2.143) and 1 Uk+1 = n! (2.144) This last equation has the solution Ant Uk |A| < 1, (2.145) [f® (a)/n!]1/(n=1)* where A is an arbitrary constant satisfying the condition given by equation (2.145). Let us now make the following definitions: 1 t = A"", B1 = (2.146) [f(m)(a)/n!]l/(n-1)' and assume that Yk = z(t) = a+Bịt+B2t2 +.... (2.147) Note that t" = An(*+1) (2.148) %3D Therefore, from yk+1 = f(yk), we have z(t") = f[z(t)]. (2.149) On substituting the expansion of equation (2.147) into this expression, we obtain the result 1 a + Bit" + B2ť²n + ...= f(a) +f (a)(Bịt+ B2ť² %3D n! 1 + B3t° + ...)" + f(a+D(a)(B¡t (2.150) (n + 1)!" + B2t? + B3t³ + ...)"+1 + ....
Yk+1 = f(a) + f^(a)(uk)", (2.143) and 1 Uk+1 = n! (2.144) This last equation has the solution Ant Uk |A| < 1, (2.145) [f® (a)/n!]1/(n=1)* where A is an arbitrary constant satisfying the condition given by equation (2.145). Let us now make the following definitions: 1 t = A"", B1 = (2.146) [f(m)(a)/n!]l/(n-1)' and assume that Yk = z(t) = a+Bịt+B2t2 +.... (2.147) Note that t" = An(*+1) (2.148) %3D Therefore, from yk+1 = f(yk), we have z(t") = f[z(t)]. (2.149) On substituting the expansion of equation (2.147) into this expression, we obtain the result 1 a + Bit" + B2ť²n + ...= f(a) +f (a)(Bịt+ B2ť² %3D n! 1 + B3t° + ...)" + f(a+D(a)(B¡t (2.150) (n + 1)!" + B2t? + B3t³ + ...)"+1 + ....
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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Chapter2: Graphical And Tabular Analysis
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
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Question
Explain this and the theorem of part 2 it is from equation2.150
![(ii) a =
f(a) and f'(a) = 0
Let f(n) (a) be the first nonzero derivative and let yk = a + uk. Therefore, in
the first approximation
Y+1 = S(a) + f (a)()",
(2.143)
Yk+1 =
and
1
Uk+1 =
n!
(2.144)
This last equation has the solution
92
Ant
Uk =
|A| < 1,
(2.145)
[fm) (a)/n!]1/(n-1)
where A is an arbitrary constant satisfying the condition given by equation
(2.145).
Let us now make the following definitions:
1
= A"",
B1
[f(m) (a)/n!]!/(n=1)
(2.146)
t =
and assume that
Yk = z(t) = a + Bịt + B2t2 + ....
(2.147)
Note that
t" = An(k+1)
(2.148)
Therefore, from yk+1 =
f(yk), we have
z(t") = f[z(t)].
(2.149)
On substituting the expansion of equation (2.147) into this expression, we
obtain the result
1
a + Bit" + Bot2n +.= f(a) + f(") (a)(Bit+ B2t?
n!
1
+ Bzt° + .)" +
(n + 1)! F*+1)(a)(Bịt (2.150)
+ Bzt? + B3t³ + ...)"+1 + ....
FIRST-ORDER DIFFERENCE EQUATIONS
69
Equating powers of t on each side allows the determination of B2, B3, etc.
For n = 2, we have
z(t?) = f[2(t)],
(2.151)
a = f(a),
f'(a) = 0,
(2.152)
and
B1 =
f"(a)'
2f"(a)
3f"(a)3*
5 f" (a)³ f"(a)
B2 = -
(2.153)
f"'(a)
B3
9f"(a) 3f"(a)ª f"(a)³'
etc.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff127b632-c7db-43de-8096-c44c67b91c3f%2F24b0def7-7691-48e5-8425-f43644e8f873%2Fblce9p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(ii) a =
f(a) and f'(a) = 0
Let f(n) (a) be the first nonzero derivative and let yk = a + uk. Therefore, in
the first approximation
Y+1 = S(a) + f (a)()",
(2.143)
Yk+1 =
and
1
Uk+1 =
n!
(2.144)
This last equation has the solution
92
Ant
Uk =
|A| < 1,
(2.145)
[fm) (a)/n!]1/(n-1)
where A is an arbitrary constant satisfying the condition given by equation
(2.145).
Let us now make the following definitions:
1
= A"",
B1
[f(m) (a)/n!]!/(n=1)
(2.146)
t =
and assume that
Yk = z(t) = a + Bịt + B2t2 + ....
(2.147)
Note that
t" = An(k+1)
(2.148)
Therefore, from yk+1 =
f(yk), we have
z(t") = f[z(t)].
(2.149)
On substituting the expansion of equation (2.147) into this expression, we
obtain the result
1
a + Bit" + Bot2n +.= f(a) + f(") (a)(Bit+ B2t?
n!
1
+ Bzt° + .)" +
(n + 1)! F*+1)(a)(Bịt (2.150)
+ Bzt? + B3t³ + ...)"+1 + ....
FIRST-ORDER DIFFERENCE EQUATIONS
69
Equating powers of t on each side allows the determination of B2, B3, etc.
For n = 2, we have
z(t?) = f[2(t)],
(2.151)
a = f(a),
f'(a) = 0,
(2.152)
and
B1 =
f"(a)'
2f"(a)
3f"(a)3*
5 f" (a)³ f"(a)
B2 = -
(2.153)
f"'(a)
B3
9f"(a) 3f"(a)ª f"(a)³'
etc.
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