Can someone please show how to derive the Newton root finding formula from a term taylor series. My main issue is I am not sure what mathematically the Newton Root finding formula actually is as I have only learned about it through my numerical methods class through MATLAB

Two Term Taylor Series: $f(x_i) + f '(x-x_i) + f ''(x_i)(1/2)(x-x_i)^2$

Thanks for the help!

  • 2
    $\begingroup$ There appears to be 3 terms in your expansion $\endgroup$ Commented Dec 10, 2016 at 17:00

2 Answers 2


We have that the 2 term Taylor polynomial is given by

$$f(x)\approx f(x_0)+f'(x_0)(x-x_0)$$

Setting $f(x)=0$, we end up with

$$ \begin{align} 0&=f(x_0)+f'(x_0)(x-x_0)\\ f'(x_0)(x-x_0)&=-f(x_0)\\ x-x_0&=-\frac{f(x_0)}{f'(x_0)}\\ x&=x_0-\frac{f(x_0)}{f'(x_0)} \end{align} $$

  • $\begingroup$ @SRW7 No problem, hope you found this useful! $\endgroup$ Commented Dec 10, 2016 at 17:17
  • $\begingroup$ Oops I'm not sure anymore, forget it. $\endgroup$
    – reuns
    Commented Dec 10, 2016 at 17:52
  • $\begingroup$ @user1952009 What? $\endgroup$ Commented Dec 10, 2016 at 18:02

Suppose $f(a)=0$, then Taylor's formula says:


for some $\xi$ between $a$ and $x$. Solving for $a$ (sort of) gives:


which, for $x\approx a$, we can (under certain conditions) expect to say that:

$$a\approx x-\frac{f(x)}{f'(x)}$$

which is, in essence, the Newton-Raphson method.

One can be a bit more careful with this argument to provide a more formal proof of the convergence of the method to $a$ under certain conditions.

  • $\begingroup$ Also is the last part of the Taylor Formula you gave the error formula? $\endgroup$
    – SRW7
    Commented Dec 10, 2016 at 17:20
  • $\begingroup$ Yep - this is so that the equation is actually a perfect equation, rather than an approximate equality. The approximation step comes from noting that $x\approx a\implies (a-x)^2\frac{f''(\xi)}{f'(x)}\approx 0$. $\endgroup$
    – πr8
    Commented Dec 10, 2016 at 17:22

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