Solve x4 x2 2x 3 0 by a Fixedpoint method and b Newtons

Solve x^4 - x^2 + 2x - 3 = 0, by a) Fixed-point method, and b) Newton\'s method by staring from x_0 = 1. For each method, do 5 steps of iteration and keep 5 decimal places in each iteration. Or You can also do the iterations in Excel, where the iteration can stop when |x_n+1 - x_n|

Solution

Using fixed point method, we don\'t get a solution. It keeps jumping between the two values, neither of which are an answer of the equation

Using newton\'s method:

Iterations

Using newton\'s method we get the answer as 1.1944

xi xi+1
1 1.5
1.5 0.774194
0.774194 1.775317
1.775317 0.515461
0.515461 1.850142
1.850142 0.462753
0.462753 1.833359
1.833359 0.474014
0.474014 1.837681
1.837681 0.471083
0.471083 1.836593
1.836593 0.471819
0.471819 1.836869
1.836869 0.471632
0.471632 1.836799
1.836799 0.47168
0.47168 1.836817
1.836817 0.471668
0.471668 1.836812
1.836812 0.471671
0.471671 1.836813
1.836813 0.47167
 Solve x^4 - x^2 + 2x - 3 = 0, by a) Fixed-point method, and b) Newton\'s method by staring from x_0 = 1. For each method, do 5 steps of iteration and keep 5 de

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