How does the geometric iteration process using $f(x)=x$ and $g(x)=a/x$ demonstrate convergence to $\sqrt{a}$ in the Babylonian method?

Conditions

• $a > 0$
• Initial guess $x_0 > 0$

Reasoning, step by step

1. Plot the red line $y=x$ and the green curve $y=a/x$ on a Cartesian coordinate system.
2. Identify the intersection point where $x = a/x$, implying $x^2 = a$ or $x = \sqrt{a}$.
3. Start with an initial value $x_n$ on the x-axis.
4. Draw a vertical line from $x_n$ up to intersect $y=x$ at height $x_n$ and $y=a/x$ at height $a/x_n$.
5. Calculate the midpoint of these two heights: $\frac{1}{2}(x_n + \frac{a}{x_n})$.
6. Project this midpoint height horizontally to find the next iterate $x_{n+1}$ on the x-axis.
7. Observe that repeated application causes $x_n$ to converge rapidly to $\sqrt{a}$.

Example

Starting from a point on the x-axis, vertical lines are drawn up to intersect both curves. The midpoint between these two intersection heights represents the value of the next term in the sequence. This height is then projected horizontally back onto the x-axis to locate the new iterate.

Common misconceptions

• Believing that the iteration converges to the average of the x-values rather than the midpoint of the y-heights.
• Assuming the method works for negative $a$ without modification (the graph would not have real intersections).

Watch the explanation

Explore next

• Why is the arithmetic mean used instead of the geometric mean in this specific graphical construction?
• What happens to the convergence speed if the initial guess $x_0$ is very far from $\sqrt{a}$?