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Before I am told, I want to clarify that I searched first, and I don't believe this to be a repost. I understand the formula in terms of how to apply it, and I've seen graphical representations and everything. I get that we are finding where the tangent line has a root, then choosing a new $f(x)$ at that point and finding the root of its tangent line, effectively closing the distance between x and the root r.

What I do not understand, is what $\frac{f(x)}{f'(x)}$ is actually doing. I know it can be used to find the root x, as it is derived from $y=mx+b$, but how is dividing $f(x)$ by its derivative getting me the root? Why does this work? My intuition is telling me (before I actually tried it) that I was getting some y value, then seeing how many times the slope goes into it; but this would give me the $x$ coordinate, wouldn't it? I can use it, but it's not clicking as to why, and I'd like to fix that so I can actually understand what is going on.

Alex G
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  • https://www.math.ubc.ca/~anstee/math104/104newtonmethod.pdf see section 2.3. i think you'll get some useful – iostream007 Apr 21 '15 at 06:42
  • Also notice that Newton's method quite often doesn't work. The idea as can be seen from the answers that for it to work well you should be quite close to the root already or the function needs to be quite nice. – DRF Apr 21 '15 at 09:17

2 Answers2

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When you are at $x_0$ which is "close" to the root, you can expand the function as a truncated Taylor series, that is to say $$f(x)=f(x_0)+(x-x_0) f'(x_0)+O\left((x-x_0)^2\right)$$ So, locally, $$f(x)\approx f(x_0)+(x-x_0) f'(x_0)=0$$ from which the update $$x=x_0-\frac{f(x_0)}{f'(x_0)}$$ and repeat until convergence.

Claude Leibovici
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To understand the method, replace the function $f(x)$ by its first order Taylor expansion

$f_L(x)=f(x_0)+f'(x_0)(x-x_0)$

at the point $x_0$ and then calculate the point where this function is zero. Then you will see from where the term $f(x)/f'(x)$ is coming.

Karl
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