F(x)=x⁴-1
f'(x)=4x³
Newton’s Method: x[n+1]=x[n]-f(x[n])/f'(x[n]); x[n+1]=x[n]-(x[n]⁴-1)/4x[n]³
x₁=3.00390625
x₂=2.26215...
x₃=1.7182...
X'=X-(X⁴-1)/4X³=X-X/4+1/4X³ is a symbolic way of writing the recursive formula, where X' represents the next iteration.
When X'≈X, -X/4+1/4X³≈0; so X/4≈1/4X³; X≈1/X³, so X⁴≈1 and X⁴-1≈0. But this is f(x)≈0. Hence Newton’s Method converges to a solution.
The rate of change is x[n+1]-x[n]=-(x[n]⁴-1)/4x[n]³=x[n]/4-1/4x[n]³ or symbolically -X/4+1/4X³.
Note that the method converges to one solution. A different x₀ will possibly converge to the solution x=-1.
Answer:
False. false false false false
9/2 that the answer ok ok
Answer
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Step-by-step explanation:
