Explain the difference between the two cases

Context: This was solved using Newton's Method. I understand that the initial guess is the importance factor here. But why is the first case converging at a quicker rate. Is it because my initial guess of 3.11 (p0) was already much closer to the root of p than 3.12? Theory would be greatly appreciated.

In [184]: T = 0.75 P = 0.3 newton(f, f_prime, 3.11, le-10, 25) n 1 2 3 4 5 6 7 8 pe 3.1100000000 0.9308540944 2.5872126976 1.4473134304 1.3237659801 1.3108041816 1.3106443183 1.3106442938 pn 0.9308540944 2.5872126976 1.4473134304 1.3237659801 1.3108041816 1.3106443183 1.3106442938 1.3106442938 f(po) -9.8578023333 0.7163573499 -6.9989463218 -0.4915013504 -0.0424583751 -0.0005110245 -0.0000000784 -0.0000000000 Approximate root was found after 8 iterations. Approximate root is 1.3106442938034883 Case E: T = 0.75 and P = 0.3 with initial guess 3.12. In [208]: T = 0.75 P = 0.3 approx = newton(f, f_prime, 3.12, le-10,25) n 1 2 3 4 5 6 7 8 9 10 11 3.1200000000 0.9079723612 3.9454083508 11.7564737279 8.7892143199 6.9124440370 5.8234222991 5.3268623228 5.2071933504 5.2003133331 5.2002910813 5.2002910810 P_n 0.9079723612 3.9454083508 11.7564737279 8.7892143199 6.9124440370 5.8234222991 5.3268623228 5.2071933504 5.2003133331 5.2002910813 5.2002910810 5.2002910810 f(po) -9.9028053333 0.7240385702 -11.4277780875 771.6458720960 222.7761320169 61.6075182932 15.0004957276 2.4592349688 0.1268942199 0.0004077737 0.0000000043 -0.0000000000 12 Approximate root was found after 12 iterations. Approximate root is 5.200291081039649 In [184]: T = 0.75 P = 0.3 newton(f, f_prime, 3.11, le-10, 25) n 1 2 3 4 5 6 7 8 pe 3.1100000000 0.9308540944 2.5872126976 1.4473134304 1.3237659801 1.3108041816 1.3106443183 1.3106442938 pn 0.9308540944 2.5872126976 1.4473134304 1.3237659801 1.3108041816 1.3106443183 1.3106442938 1.3106442938 f(po) -9.8578023333 0.7163573499 -6.9989463218 -0.4915013504 -0.0424583751 -0.0005110245 -0.0000000784 -0.0000000000 Approximate root was found after 8 iterations. Approximate root is 1.3106442938034883 Case E: T = 0.75 and P = 0.3 with initial guess 3.12. In [208]: T = 0.75 P = 0.3 approx = newton(f, f_prime, 3.12, le-10,25) n 1 2 3 4 5 6 7 8 9 10 11 3.1200000000 0.9079723612 3.9454083508 11.7564737279 8.7892143199 6.9124440370 5.8234222991 5.3268623228 5.2071933504 5.2003133331 5.2002910813 5.2002910810 P_n 0.9079723612 3.9454083508 11.7564737279 8.7892143199 6.9124440370 5.8234222991 5.3268623228 5.2071933504 5.2003133331 5.2002910813 5.2002910810 5.2002910810 f(po) -9.9028053333 0.7240385702 -11.4277780875 771.6458720960 222.7761320169 61.6075182932 15.0004957276 2.4592349688 0.1268942199 0.0004077737 0.0000000043 -0.0000000000 12 Approximate root was found after 12 iterations. Approximate root is 5.200291081039649