Explain the determaine (ii) a = f(a) and f'(a) = 0Let f(n) (a) be the first nonzero derivative and let yk = a + uk. Therefore, inthe first approximation1Yk+1 = f(a) +f(n) (a)(uk)";n!"(2.143)and1Uk+1 =п!¡ (m)(a)(uk)".(2.144)This last equation has the solutionAn'|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:1t = An*, B1 =(2.146)[f(m) (a)/n!]1/(n-1)'and assume thatYk = z(t) = a + Bịt + B2t2 + · . .(2.147)Note thatAn(k+1)(2.148)tn =Therefore, from yk+1 =f(yk), we havez(t") = f[z(t)].(2.149)On substituting the expansion of equation (2.147) into this expression, weobtain the result1: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 + · · · .

Name: Explain the determaine (ii) a = f(a) and f'(a) = 0Let f(n) (a) be the
Uploaded: 2024-03-20T07:06:41.000Z
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Description: Explain the determaine (ii) a = f(a) and f'(a) = 0Let f(n) (a) be the first nonzero derivative and let yk = a + uk. Therefore, inthe first approximation1Yk+1 = f(a) +f(n) (a)(uk)";n!"(2.143)and1Uk+1 =п!¡ (m)(a)(uk)".(2.144)This last equation has the