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In lecture, we were told that to find $\sqrt[3]{a}$, we use Newton's method as follows: $$ \begin{align} f(x) &= x^3 - a\\ f'(x) &= 3x^2\\ x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)}\\ &= x_n - \frac{{x_n}^3-a}{3{x_n}^2}\\ &=\frac{2}{3}x_n + \frac{a}{3{x_n}^2} \end{align} $$

And for approximating $\frac{1}{a}$,

$$ \begin{align} f(x) &= a - \frac{1}{x}\\ f'(x) &= \frac{1}{x^2}\\ x_{n+1} &= x_n - \frac{x_n}{x_n}\\ &= x_n - (a - \frac{1}{x_n})x^2\\ &=2x_n - a{x_n}^2 \end{align} $$ $$ \\ \\ $$ However, I didn't follow why $f(x) =x^3 - a$ instead of $f(x)=\sqrt[3]{x}$ for the first case and why $f(x)=a - \frac{1}{x}$ instead of $f(x)= \frac{1}{x}$ in the second case.

In each case, the teacher mentioned that the value we're trying to solve for can be realized as the root of $f(x).$ I'm not sure I know what that means in relation to finding a proper f(x).

From what I understand, we are trying to find successively closer $x_{n+1}$ values to the root (where the function is zero) by finding the x intercept of the tangent line of our function at $x_n$ and then repeating that process.

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  • $\begingroup$ incidentally, your iteration appears correct. It's just the interpretation of the term "root" that slipped past you. The good news is you can do the hard part, what you're missing is the easiest part. $\endgroup$ Jun 13, 2013 at 21:39

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A root for $f(x)$ is simply a value $x_o$ such that $f(x_o)=0$. For the functions you listed each root is the value you seek.

For $f(x)=a-x^3$ if $f(x_o)=a-x_o^3=0$ then $x_o^3=a$ hence $x_o = \sqrt[3]{a}$.

For $f(x)=a-1/x$ if $f(x_o)=a-1/x_o=0$ then $1/x_o=a$ hence $x_o = 1/a$.

The reason not to do what you say is that these work.

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  • $\begingroup$ Ah, it's easier for me to see by working backwards from $x_0$ $\endgroup$
    – maogenc
    Jun 13, 2013 at 22:02
  • $\begingroup$ @maogenc sorry for using $x_o$, I should really have used some other nuetral letter or perhaps $x_*$ to indicate the root is the fixed point of a successful iteration (it's possible for Newton's Method to fail by cycling around the answer etc... but I don't believe it fails your cases) $\endgroup$ Jun 14, 2013 at 1:50

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