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

numerical methodsroots

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?

Best Answer

(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. enter image description here

(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
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