Question

find the real root of the equation X3 = X2+1
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Answer 1

f(x) = x^3 + x^2 - 1 = 0.

x^2 (x + 1) = 1.

x^2 = 1 / (x + 1).

x = sqrt (1 / (x+1)). In general, g(x) = Xn+1 = sqrt (1 / (Xn + 1))

Let us take an initial guess of 0. Xo = 0

X1 = sqrt ( 1 / 1) = 1

X2 = sqrt (1 / 2) = sqrt (0.5) = 0.707

X3 = sqrt (1 / 1.707) = 0.765

X4 = sqrt (1 / 1.765) = 0.752

X5 = sqrt (1 / 1.752) = 0.755

Thus, the root of the equation is approximately 0.755.