how can we sketch the graph of this iteration xn+1=xn-8*f'(xn) in matlab code?

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biratu lemessa
biratu lemessa 2023-5-4
回答： Walter Roberson 2023-5-4
f(x)=x^2;
x0=0.5;
tolerance =0.0001
xn+1=xn-8*f'(xn)

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回答（2 个）

Cris LaPierre
Cris LaPierre 2023-5-4
Have you tried the plot command?
See Ch 9 of MATLAB Onramp.
Walter Roberson
Walter Roberson 2023-5-4
consider
f(x)=x^2;
x(n+1) = x(n) -8*f'(x(n))
But when f(x) = x^2 then f'(x) = 2*x so x(n+1) = x(n) - 8*2*x(n) = -15*x(n)
and so X(N) = (-15)^N * X0^N
When X0 > 1/15 then X0^N does not shrink faster than the 15^N grows so this equation is divergent for X0 as large as 1/2 and will be approximately (-8)^N

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