Suppose that for $f(x)$, $\frac{f(x)}{f'(x)}=3x$. If I wanted to find a root for $f$ using the Newton-Raphson method I have the formula $x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} = x_{n} - 3x_{n} = -2x_{n}$ but trying it out it just oscillates between numbers and doesn't give a concrete result. Why does this happen?
2026-03-28 00:28:40.1774657720
Why doesn't Newton-Raphson work for these type of functions?
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The "correction term" you've given can be treated as a differential equation which yields $$ \frac{y}{dy/dx} \ \ = \ \ 3x \ \ \rightarrow \ \ \frac{dy}{y} \ \ = \ \ \frac13 \ \frac{dx}{x} \ \ \Rightarrow \ \ \ln y \ \ = \ \ \frac13 \ln x \ \ + \ \ C \ \ \Rightarrow \ \ y \ \ = \ \ A·x^{1/3} \ \ . $$
So your example is some multiple of a cube-root function, which has very shallow slope over most of its curve. Since the Newton-Raphson method uses $ \ x-$intercepts of tangent lines to the function curve as "iterates" to the zero, unless our choice for $ \ x_0 \ $ is "quite close" to the zero of this function, the intercept of the tangent line at $ \ x_0 \ $ will be very far from the zero we are seeking. If $ \ x_1 \ $ lies on the opposite side of that zero from $ \ x_0 \ \ , \ $ we are now on the part of the curve with slope of opposite sign. This will set up an "oscillation" around the zero for the successive iterations, with no likelihood of convergence. (Here we have a "geometrically increasing" oscillation, as described in the comments above.)
In the "correction terms", we really want to see $ \ \frac{f(x)}{f'(x)} \ = \ C·x_n \ \ $ with $ \ |C| \ << \ 1 \ \ $ in order to have rapid convergence.
Newton-Raphson is not a "universal" method and quite a lot has been written on the various ways it can fail to converge. As is apparent, there are types of functions and "function behavior" in the neighborhood of zeroes for which this method cannot be recommended.