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We want to compute the root of $$f(x) = x^3-2x-5$$

Since $f(1.5)<0, f(2.5)>0$ the root must be within the interval $[1.5,2.5]$.

1) An intuitive iterative function would be $\phi(x) = 0.5 \cdot (x^3-5)$, that defines the iterative procedure:

$$x_{n+1} = 0.5 \cdot (x^3_n - 5)$$ but, it is not right because within the interval $[1.5, 2.5]$ there is $|\phi'(x)|>1$.

2) Another iterative function is $\phi(x) = (2x_n + 5)^{1/3}$, that defines the iterative procedure:

$$x_{n+1} = (2x_n+5)^{1/3}$$ and is ok because $1/6 < \phi'(x) < 0.15$. The procedure converges.

My questions:
- I don't understand what are the passages that bring to consider those iterative functions $\phi(x)$?

please, can you hep me? Thanks!

JB-Franco
  • 852

2 Answers2

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There is only one sign change in the coefficient sequence, so that Descartes rule predicts exactly one positive real root.

A classical method to approximate positive roots is to find a form of the equation where all coefficients are positive $$ x^3=2x+5 $$ and successively solve for the side with the higher degree $$ x_{n+1}=\sqrt[3]{2x_n+5}. $$ This is what you verified as being successful. \begin{align} x_{0}&=2.000000000000\\ x_{1}&=2.080083823052\\ x_{2}&=2.092350677798\\ x_{3}&=2.094216996013\\ x_{4}&=2.094500652195\\ x_{5}&=2.094543757533\\ \end{align}

Another approach is to use that you already found that the root is close to $2$ and perform polynomial division by $x-2$, $$ 0=(x-2)(x^2+2x+2)-1 $$ and derive the iteration $$ x_{n+1}=2+\frac1{(x_n+1)^2+1} $$ \begin{align} x_{0}&=2.000000000000\\ x_{1}&=2.100000000000\\ x_{2}&=2.094250706880\\ x_{3}&=2.094568125667\\ x_{4}&=2.094550560622\\ x_{5}&=2.094551532497\\ \end{align}

A more systematic approach would be to use the simplified Newton method starting at the guessed location of the root $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(2)}=x_n-\frac{x^3-2x-5}{10} $$ \begin{align} x_{0}&=2.000000000000\\ x_{1}&=2.100000000000\\ x_{2}&=2.093900000000\\ x_{3}&=2.094626880398\\ x_{4}&=2.094542720862\\ x_{5}&=2.094552498993\\ \end{align} or the full Newton method, using some modification $f(x)=x^2-2-5x^{-1}$ of the equation for variety $$ x_{n+1}=x_n-\frac{f(x)}{f'(x)} =x_n-\frac{x_n^2-2-5x_n^{-1}}{2x_n+5x_n^{-2}} =x_n\frac{x_n^3+2x_n+10}{2x_n^3+5} $$ \begin{align} x_{0}&=2.000000000000\\ x_{1}&=2.095238095238\\ x_{2}&=2.094551521901\\ x_{3}&=2.094551481542\\ x_{4}&=2.094551481542\\ x_{5}&=2.094551481542\\ \end{align}

Lutz Lehmann
  • 126,666
1

Another iterative (polynomial) function.

Since $f'(x)=3x^2-2$, it follows that both $f$ and $f'$ are increasing in $[1.5,2.5]$ and therefore $$f([1.5,2.5])=[f(1.5),f(2.5])]=[-4.625,5.625]$$ and $$f'([1.5,2.5])=[f'(1.5),f'(2.5])]=[4.75, 16.75].$$

Now consider the function $$\phi(x):=x-\frac{f(x)}{M}$$ with $M>0$ large enough ($M=10$ works) such that $$\phi([1.5,2.5])\subset [1.5,2.5]\qquad\text{and}\qquad\max_{x\in [1.5,2.5]} |\phi'(x)|<1.$$ Then the iterative method $x_{n+1}=\phi(x_n)$ for $n\geq 0$ with starting point $x_0\in [1.5,2.5]$ finds the unique fixed point of $\phi$ in $[1.5,2.5]$ which is the unique root of $f$ in $[1.5,2.5]$.

Robert Z
  • 145,942