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Let r be a root of order 3 $f(r)=f'(r)=f''(r)=0, f'''(r)\ne0$ and let en=xn-r, show that $$e_{n+1}=\dfrac{2}{3}e_n$$ Also, show that in case of multiplicity of 2, the modified newtons method $$x_{n+1}=x{n}-2f(x_n)/f'(x_n)$$ converges quadratically to the root.

I know how to prove that m-1/m convergence rate for m multiplicity as follows: Let $r$ be a root of multiplicity $m$. Then one can extract $m$ linear factors $(x-r)$ from $f$, so that $f(x)=(x-r)^mg(x)$, $g(r)\ne 0$, $g$ at least differentiable. Then $$f'(x)=m(x-r)^{m-1}g(x)+(x-r)^mg'(x)$$ and the Newton step gives $$ x_{n+1}-r=x_n-r-\frac{(x_n-r)^mg(x_n)}{m(x_n-r)^{m-1}g(x_n)+(x_n-r)^mg'(x_n)} \\~\\ =\frac{(m-1)g(x_n)+(x_n-r)g'(x_n)}{mg(x_n)+(x_n-r)g'(x_n)}(x_n-r) $$ which implies \begin{align} e_{n+1} &=\frac{(m-1)g(r)+e_ng'(r)+O(e_n^2)}{mg(r)+e_ng'(r)+O(e_n^2)}e_n \\[1em] &=\frac{m-1}{m}e_n+\frac{g'(r)}{mg(r)}e_n^2+O(e_n^3) \end{align}

However, is there any easier way to show this? question asks me to use f(xn) and f'(xn) at r taylor expansions. I am completely confused. For the second part of the question, it asks me to use f(r+en) and f'(r+en) to prove it. All I got was $f(r+en)=f(xn)=f(r)+f'(r)*en+f''(r)/2*en^2$ then take derivative of that to get f'(xn) and plug into the modifed method but with nothing to show after.Please explain if you can thanks.

Aniruddha Deshmukh
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james black
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1 Answers1

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Yes, you can do that, you only need to apply the Taylor expansion also to the derivative, $$ f'(r+e_n)=f'(r)+f''(r)e_n+\frac12f'''(r)e_n^2+... $$ You need to use a degree of the Taylor expansion that includes at least one non-zero term.

Then insert both into the Newton step formula and apply division rules for Taylor series.

Lutz Lehmann
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  • but I am left with some en*f''(x) term i just assume it to be O(en)? – james black Feb 19 '18 at 07:41
  • Yes if you want to compute the constant for $e_n\to 0$. It may also be a hint of a calculation error. What is $x$? If $f''(r)=0$ then $f''(x_n)=f'''(r)e_n+f^{(4)}(r)e_n^2+...$ so that your term is $O(e_n^2)$. – Lutz Lehmann Feb 19 '18 at 07:46
  • got it also out of curiisoity is newtons method derived from first three instead of the usual two terms the halley method? so thehalley method is essntially newton? – james black Feb 19 '18 at 07:48
  • Yes. Depending on how you use these three terms you get the original (with square roots) or current Halley's method or Laguerre's method. – Lutz Lehmann Feb 19 '18 at 08:07