(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 Yk+1 = f(a)+ f") (a)(uk)", (2.143) and Uk+1 = n! s() (a)(uk)" . (2.144) This last equation has the solution Ank [f(m) (a)/n!]1/(n-1) |A| < 1, (2.145) Uk where A is an arbitrary constant satisfying the condition given by equation (2.145). Let us now make the following definitions: 1 = A"*, B1 = (2.146) [S(m) (a)/n!]l/(n=1) ' and assume that Yk = z(t) = a + B1t+B2t2 + · · ·. (2.147) || Note that t" = An(*+1) (2.148) || Therefore, from Yk+1 = f(yk), we have z(t") = f[z(t)]. (2.149)
(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 Yk+1 = f(a)+ f") (a)(uk)", (2.143) and Uk+1 = n! s() (a)(uk)" . (2.144) This last equation has the solution Ank [f(m) (a)/n!]1/(n-1) |A| < 1, (2.145) Uk where A is an arbitrary constant satisfying the condition given by equation (2.145). Let us now make the following definitions: 1 = A"*, B1 = (2.146) [S(m) (a)/n!]l/(n=1) ' and assume that Yk = z(t) = a + B1t+B2t2 + · · ·. (2.147) || Note that t" = An(*+1) (2.148) || Therefore, from Yk+1 = f(yk), we have z(t") = f[z(t)]. (2.149)
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Explain the determaine
![(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
1
Yk+1 = f(a) +
f(n) (a)(uk)";
n!"
(2.143)
and
1
Uk+1 =
п!
¡ (m)(a)(uk)".
(2.144)
This last equation has the solution
An'
|A| < 1,
(2.145)
Uk
[f(m) (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 = An*, B1 =
(2.146)
[f(m) (a)/n!]1/(n-1)'
and assume that
Yk = z(t) = a + Bịt + B2t2 + · . .
(2.147)
Note that
An(k+1)
(2.148)
tn =
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
:n)
a + Bit" + B2t²n + ...= f(a)+fm) (a)(Bit+ B2ť?
n!
1
+ B3t + ..·)" +
if(a)(Bit (2.150)
:(n+1)
(n + 1)!
+ B2t? + B3t +..
· )n+1 + · · · .](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3d661a9-09b0-4ac1-a22e-91bdec7a85f9%2F87a1b6eb-6562-4581-b2d1-e4d176e957ec%2Fdmsenji_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
1
Yk+1 = f(a) +
f(n) (a)(uk)";
n!"
(2.143)
and
1
Uk+1 =
п!
¡ (m)(a)(uk)".
(2.144)
This last equation has the solution
An'
|A| < 1,
(2.145)
Uk
[f(m) (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 = An*, B1 =
(2.146)
[f(m) (a)/n!]1/(n-1)'
and assume that
Yk = z(t) = a + Bịt + B2t2 + · . .
(2.147)
Note that
An(k+1)
(2.148)
tn =
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
:n)
a + Bit" + B2t²n + ...= f(a)+fm) (a)(Bit+ B2ť?
n!
1
+ B3t + ..·)" +
if(a)(Bit (2.150)
:(n+1)
(n + 1)!
+ B2t? + B3t +..
· )n+1 + · · · .
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