f(x)=x4−x3+x−1
f′(x)=4x3−3x2+1xn+1=xn−f′(xn)f(xn) Initial solution x0=−0.5
n012345678910xn−0.5−5.75−4.26333436−3.15575240−2.33776871−1.74669885−1.34320270−1.10675944−1.01421870−1.00029532−1.00000013f(xn)−1.31251276.48828125402.49581545126.4488333339.3065189311.890741913.335309050.749348730.087146130.001772710.00000078xn+1−5.75−4.26333436−3.15575240−2.33776871−1.74669885−1.34320270−1.10675944−1.01421870−1.00029532−1.00000013−1.00000000
Δ=∣xn+1−xn∣
Δ=∣−5.75−(−0.5)∣=5.25
Δ=∣−4.26333436−(−5.75)∣=1.48666564
Δ=∣−3.15575240−(−4.26333436)∣=1.10758196
Δ=∣−2.33776871−(−3.15575240)∣=0.81798369
Δ=∣−1.74669885−(−2.33776871)∣=0.59106986
Δ=∣−1.34320270−(−1.74669885)∣=0.40349615
Δ=∣−1.10675944−(−1.34320270)∣=0.23644326
Δ=∣−1.01421870−(−1.10675944)∣=0.09254074
Δ=∣−1.00029532−(−1.01421870)∣=0.01392338
Δ=∣−1.00000013−(−1.00029532)∣=0.00029519
Δ=∣−1.00000000−(−1.00000013)∣=0.00000013
The root is −1.00000
It needs 10 iterations for this much accuracy.
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