Finding the positive root of $x^3 +x^2 =0.1$ by numerical methods.

Question

The positive root of $x^3 +x^2 =0.1$ is denoted to be $A$.

$(a)$ Find the first approximation to $A$ by linear interpolation on the interval $(0,1)$

For this, I got $x_1 =0.05$

$(b)$ Indicate why linear interpolation does not give a good approximation to $A$.

All I think of for this is that $x_1$ by linear interpolation greatly underestimates $A$ to a large extent. But wouldn't numerical methods like Newton-Raphson, Linear Interpolation and iteration $x_{n+1}=F(x_n)$ all give bad first approximations?

$(c)$ Find an alternative first approximation to $A$ by using the fact that if $x$ is small then $x^3$ is negligible compared with $x^2$

So, for this am I supposed to use Newton-Raphson with $x_1=0$ since $x$ is small?

Answer 1

(a) Correct. (b) Yes. It is also bad because $1$ is far away from the root, as can be seen comparing the function values. And the function is very non-linear around $x=0$, making linear approximations relatively bad.

(c) No, you are supposed to solve $x^2=0.1$, disregarding the $x^3$ term. Or to put it into inequalities, as $0<x<1$ you also have $$ x^2\le 0.1\le 2x^2\implies \sqrt{0.05}\le x \le \sqrt{0.1}. $$ Then iterate further, for instance using $x_{k+1}=\sqrt{\frac{0.1}{1+x_k}}$.

k    x[k]
------------------
1   0.316227766017
2   0.275635071544
3   0.279986295554
4   0.279509993492
5   0.279562012937
6   0.279556330208
7   0.279556950986
8   0.279556883172
9   0.27955689058

Answer 2

Just for your curiosity.

We can approximate functions using Padé approximants much better than with Taylor series (remember that Newton method is equivalent to an $O(x^2)$ Taylor expansion).

Since we know how to solve easily quadratic equations, let us consider the $[2,2]$ Padé approximant. It will be $$x^3+x^2 \sim\frac{x^2}{x^2-x+1}$$ If you develop the rhs of the above as a Taylor series, you would get $x^2+x^3-x^5+O\left(x^6\right)$ (pretty close, isn't it ?).

So, for small values of $a$, the approximate solution of $x^3+x^2=a$ is given by the solution of $(1-a) x^2+a x-a=0$ that is to say $$x_\pm=\frac{a\pm\sqrt{4 a-3 a^2}}{2 (a-1)}$$ So, for $a=\frac 1 {10}$, the estimate would be $x\approx 0.282376$ which is quite close to the "exact" solution $x\approx 0.279557$

Answer 3

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• Start with the iteration $$ x_{n} = \root{0.1 - x_{n}^{3}}\,,\qquad x_{0} = \root{0.1} \approx 0.3162 $$ It seems to be that $\ds{10}$ iterations are enough. Namely, $$ x_{10} \approx 0.279564\quad\mbox{and}\quad\mrm{f}\pars{x_{10}} \approx 5.43129 \times 10^{-6} $$ where $\bbox[10px,#ffd,border: 1px groove navy]{\ds{\mrm{f}\pars{x} \equiv x^{3} + x^{2} - 0.1}}$.
• In addition, you can refine the above result by means of a Newton-Rapson Iteration: $$ y_{n} = y_{n - 1} - {y_{n - 1}^{3} + y_{n - 1}^{2} - 0.1 \over 3y_{n - 1}^{2} + 2y_{n - 1}}\,,\qquad y_{0} = x_{10} \approx 0.279564 $$ With about three iterations, I'll find $$ y_{3} \approx \bbx{\large 0.279957} \implies \mrm{f}\pars{y_{3}} \approx 2.77556 \times10^{-17} $$ \begin{align} \end{align}