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.
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.