The Newton's method

This method consists of calculating a certain value that is more or less known, but which we would like to approximate more precisely. The method is done by successive iteration.

Let $ f $ be a convex function (resp. concave) and strictly increasing (resp. decreasing) and positive on a interval $ I $ and $ a_0 $ a real number belonging to $ I $ such as $ f(a_0) > 0 $ (resp. $ f(a_0) < 0 $).

The method consists of determining the real number $ x_0 \in I $, such as: $ f(x_0) = 0 $

Introduction of the Newton's method by successive iteration

In a general way, we will determine the value of $ x_0 $ for which $ f $ is worth $ 0 $ on $ I $.

And this value is worth:

$$ x_0 = lim_{n \to +\infty} \enspace (a_n) $$

With $ (a_n)_{n \in \mathbb{N}} $ a recurring sequence such as:

$$ a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} $$

Demonstration

We know that this function will cancel at some point, but we don't know yet for what value $ x_0 $.

We have represented the tangent to this curve at point $ x = a_0 $, and corresponding to the following equation:

$$ T_{a_0}(x) = f'(a_0)(x - a_0) + f(a_0)$$

We will thus try to find out when this function will cancel itself. We do have:

$$ T_{a_0}(x) = 0 $$

$$ f'(a_0)(x - a_0) + f(a_0) = 0 $$

$$ f'(a_0).x - f'(a_0).a_0 + f(a_0) = 0 $$

$$ f'(a_0).x = f'(a_0).a_0 - f(a_0) $$

At this stage, we know that $ f'(a_0) > 0 $, then we can divide by $ f'(a_0) $.

$$ x = \frac{f'(a_0).a_0 - f(a_0)}{f'(a_0)} $$

$$ x = a_1 = a_0 - \frac{ f(a_0)}{f'(a_0)} $$

If we continue and do the same thing, this time not starting from $ a_1 $, we will obtain:

$$ a_2 = a_1 - \frac{ f(a_1)}{f'(a_1)} $$

We then obtain a recurring sequence $ (a_n)_{n \in \mathbb{N}} $ such as:

$$ \enspace a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} $$

The more successive iterations we establish, the more we will tend towards the desired value, namely:

$$ x_0 = lim_{n \to +\infty} \enspace (a_n) $$

With $ (a_n)_{n \in \mathbb{N}} $ a recurring sequence such as:

$$ a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} $$

We can repeat this same reasoning for a concave and/or decreasing and/or negative function, and adapt a such process according to the case.

Example

We will try to find an approximate value of $ \sqrt{2} $ by this method.

Let $ f $ be a function such as $ x_0 = \sqrt{2}$ be a solution for$ f(x_0) = 0 $. Let's start from this hypothesis and try to determine this function.

$$ x_0 = \sqrt{2} $$

$$ x_0^2 = 2 $$

$$ x_0^2 - 2 = 0 $$

We will then study the function $ f $ defined by:

$$ f(x) = x^2 - 2 $$

And thus find an approximate value of $ x_0 $ for which $ f(x_0) = 0 $.

We are definitely in the case of a convex function and strictly increasing function, we can then apply the method.

This involves computing the different values of the following:

$$ a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} $$

With $ f(x) = x^2 - 2 $ and its derivative function $ f'(x) = 2x $.

So,

$$ a_{n + 1} = a_n - \frac{a_n^2 - 2}{2a_n} $$

Here are the results of the calculation with different value for $ a_0 $: