Cannot understand the straight line equation in Newton Raphson method.

Question

I'm currently trying to gain an intuitive understanding of the Newton Raphson method but have reached a hurdle I seem unable to jump at the moment:

Here's where I am so far:

1. We have a function $f(x)$ that we want to approximate the root of.
2. We make a guess, $g$.
3. By drawing a tangent line at $f(g)$, we have the coordinate $(g, f(g))$
4. The equation of this line in the form $y = mx + c$ is $y = f'(g) \textbf{(x -g)} + \textbf{f(g)}$

The bold text is where I'm struggling! Why is the $x$ value $(x-g)$ and why is the constant $f(g)$? The tangent line doesn't go through the y-axis there. See picture here, courtesy of Brilliant: Tangent Line

Thanks

Answer 1

• The tangent line will go through the $y$-axis if you extend it.

• Let's study $\hat{f}(x) = f'(g) (x-g)+f(g)$ and check that it passes through $(g,f(g))$.

We evaluate $f'(g)(x-g)+f(g)$ at $g$ and have $$f'(g)(g-g)+f(g)=f(g)$$

also $y= f'(g)(x-g)+f(g)=f'(g)x +(f(g)-gf'(g))$ is a straight line with slope $f'(g)$ and intercept $f(g)-gf'(g)$.

Answer 2

This is just the point-slope form for a linear equation. You have the point $(g,f(g))$ that you want the line to go through. The derivative is the slope of the tangent line at that point. The point-slope form for a point $(x_1,y_1)$ and a slope of $m$ is $y-y_1=m(x-x_1)$. Now we substitute in $x_1=g, y_1=f(g), m=f'(g)$ and you get the equation you present.

Answer 3

According to the Taylor series expansion, near $x_0$, for a regular $f(x)$ we have

$$ f(x) = f(x_0)+f'(x_0)(x-x_0) + O(|x-x_0|^2) $$

now if $x = x_1$ is such that $f(x_1)\approx 0$ we can make

$$ 0 = f(x_0)+f'(x_0)(x_1-x_0)\Rightarrow x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} $$

or recurrently

$$ x_{x+1}=x_k- \frac{f(x_k)}{f'(x_k)} $$

Now this process guide us to a root for $f(x)$ near $x_0$ if

$$ \phi(x) = x-\frac{f(x)}{f'(x)} $$

is a contraction or if $|\phi'(x_0)| < 1$