24. Newton's Method, Rolle's Theorem and Mean Value Theorem

Dated: 30-10-2024

Newton's Method for Approximating Solutions for \(f(x) = 0\)

Let us first describe what does \(f(x) = 0\) even means.
It means we are looking for those values of \(x\) where \(f(x) = y = 0\).
These are the values where the graph of function¹ crosses the x axis.

They way it works is as follows:

Choose an arbitrary \(x_1\).
Draw the tangent to \(f(x_1)\)
If the tangent is not parallel to x axis i.e. \(f^{\prime}(x_1) \ne 0\) then then tangent line intercepts at x axis at \((x_2, 0)\).
Select \(x_2\) and repeat from Step 2 until the new \(x\) values slowly seem to approach a certain value.

We start with \(x_1\).
To draw tangent at \(f(x_1)\), we are going to use the definition of slope.

\[m = f^{\prime}(x_1)\]

\[run = x_2 - x_1\]

\[rise = 0 - f(x_1)\]

\[\because m = \frac{rise}{run}\]

\[f^{\prime}(x_1) = \frac{0 - f(x_1)}{x_2 - x_1}\]

\[f^{\prime}(x_1) \cdot (x_2 - x_1) = -f(x_1)\]

\[x_2 - x_1 = \frac{-f(x_1)}{f^{\prime}(x_1)}\]

\[x_2 = x_1 - \frac{f(x_1)}{f^{\prime}(x_1)}\]

Then we can generalize the equation from here

\[x_{n+1} = x_n - \frac{f(x_n)}{f^{\prime}(x_n)}\]

Limitations

Does not always work.
It might involve division by \(0\). Hence rendering us unable to continue the process. This case can occur if he slope is \(0\) at any point.
Sometimes, the approximations do not converge to a solution.

Rolle's Theorem

If \(f(x)\) is differentiable over an interval² \((a, b)\) and continous³ over the interval² \([a, b]\) then there exists \(c \in \{x \in \mathbb{R}: a < x < b\}\) such that \(f^{\prime}(c) = 0\).

General Equation

\[\because \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}\]

\[\frac{y - f(a)}{x - a} = \frac{f(b) - f(a)}{b - a}\]

\[y - f(a) = \frac{f(b) - f(a)}{b - a} \cdot (x - a)\]

\[y = \frac{f(b) - f(a)}{b - a} \cdot (x - a) + f(a)\]

Mean Value Theorem

If the conditions are right, the secant line will have same slope as the tangent line.

If \(f(x)\) is differentiable over an interval² \((a, b)\) and continous³ over the interval² \([a, b]\) then there exists \(c \in \{x \in \mathbb{R}: a < x < b\}\) such that \(f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}\).

Proof

We can define a function¹ \(v(x)\) which represents the distance between the curve and the secant line.
From Rolle's Theorem

\[v(x) =f(x) - \left(\frac{f(b) - f(a)}{b - a} \cdot (x - a) + f(a)\right)\]

\[v^{\prime}(x) = f^{\prime}(x) - \left(\frac{f(b) - f(a)}{b - a}\right)\]

Since, \(v(a) = v(b) = 0\), it satisfies rolle's theorem and there is a point \(c\) where \(v^{\prime}(c) = 0\)

\[v^{\prime}(c) = f^{\prime}(c) - \left(\frac{f(b) - f(a)}{b - a}\right)\]

\[0 = f^{\prime}(c) - \left(\frac{f(b) - f(a)}{b - a}\right)\]

\[f^{\prime}(c) = \left(\frac{f(b) - f(a)}{b - a}\right)\]