Cfrgtkky

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?

(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
$$
x^2le 0.1le 2x^2implies sqrt{0.05}le x le sqrt{0.1}.
$$
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

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
$$x^3+x^2 simfrac{x^2}{x^2-x+1}$$ If you develop the rhs of the above as a Taylor series, you would get $x^2+x^3-x^5+Oleft(x^6right)$ (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
$$x_pm=frac{apmsqrt{4 a-3 a^2}}{2 (a-1)}$$ So, for $a=frac 1 {10}$, the estimate would be $xapprox 0.282376$ which is quite close to the "exact" solution $xapprox 0.279557$

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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, $$
  x_{10} approx 0.279564quadmbox{and}quadmrm{f}pars{x_{10}} approx 5.43129 times 10^{-6}
  $$
  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}