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: \( f(x_0) = 0 \)
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:
$$ a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} $$
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. \( f(a_0) < 0 \)).
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.
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 \( a_0 \):
Go to the top of the page
