Newton's Approximation For Pi

I watched an interesting YouTube video about how Newton approximated \( \pi \) and I thought it would be an interesting exercise to get Gemini to access the video and recreate the steps outlined. Once Gemini had created its response, I then got it to create a HTML version of the information for easy pasting into Blogger. Here is what it came up with and it's really quite impressive:

Based on the video provided, specifically the section starting around [00:06:50], here is the step-by-step derivation of Newton’s approximation of $\pi$.

Newton's method combines two different ways of calculating the area of a specific portion of a circle: using calculus (integration of an infinite series) and geometry. By equating these two results, he solved for $\pi$.

Step 1: The Equation of the Circle

Newton started with the equation of a unit circle centered at the origin:

$$x^2 + y^2 = 1$$

Solving for $y$ gives the function for the upper semicircle:

$$y = \sqrt{1 - x^2} = (1 - x^2)^{\frac{1}{2}}$$

Newton wanted to find the area under this curve from $x = 0$ to $x = \frac{1}{2}$. This can be expressed as the definite integral:

$$\text{Area} = \int_{0}^{\frac{1}{2}} (1 - x^2)^{\frac{1}{2}} \, dx$$

Step 2: Binomial Expansion

Since calculus was in its infancy, Newton applied his generalized binomial theorem to expand $(1 - x^2)^{\frac{1}{2}}$ into an infinite series. The general form for $(1+u)^n$ is:

$$(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots$$

By substituting $u = -x^2$ and $n = \frac{1}{2}$, he derived the series for the circle's equation:

$$(1 - x^2)^{\frac{1}{2}} = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{16}x^6 - \frac{5}{128}x^8 - \dots$$

Step 3: Term-by-Term Integration

Newton then found the area (fluxion) by integrating the series term by term:

$$\int \left( 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{16}x^6 - \dots \right) \, dx$$

This yields the algebraic formula for the area:

$$\text{Area}(x) = x - \frac{1}{6}x^3 - \frac{1}{40}x^5 - \frac{1}{112}x^7 - \frac{5}{1152}x^9 - \dots$$

Step 4: Calculating the Area for $x = \frac{1}{2}$

Newton improved efficiency by choosing the limit $x = \frac{1}{2}$ (instead of 1), which makes the powers of $x$ shrink very rapidly. Substituting $x = \frac{1}{2}$ into the series:

$$\text{Area} \approx \frac{1}{2} - \frac{1}{6}\left(\frac{1}{2}\right)^3 - \frac{1}{40}\left(\frac{1}{2}\right)^5 - \frac{1}{112}\left(\frac{1}{2}\right)^7 - \dots$$

Simplifying the fractions:

$$\text{Area} \approx \frac{1}{2} - \frac{1}{48} - \frac{1}{1280} - \frac{1}{14336} - \dots$$

This sum gives a numerical value for the area.

Step 5: Equating to Geometry

Screenshot from video

Geometrically, the area under the unit circle from $x=0$ to $x=\frac{1}{2}$ corresponds to a slice of the circle. As described in the video, this region consists of two shapes:

1. A triangle with base $\frac{1}{2}$ and height $\frac{\sqrt{3}}{2}$ (since $y = \sqrt{1 - (0.5)^2} = \sqrt{0.75} = \frac{\sqrt{3}}{2}$).
2. A circular sector of $30^{\circ}$ (or $\frac{\pi}{6}$ radians).

The total geometric area is:

$$\text{Area} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{sector}}$$ $$\text{Area} = \left( \frac{1}{2} \cdot \text{base} \cdot \text{height} \right) + \left( \frac{\theta}{2\pi} \cdot \pi r^2 \right)$$ $$\text{Area} = \left( \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{\pi/6}{2\pi} \cdot \pi (1)^2 \right)$$ $$\text{Area} = \frac{\sqrt{3}}{8} + \frac{\pi}{12}$$

Step 6: Solving for $\pi$

Finally, Newton equated the Algebraic Area (from Step 4) with the Geometric Area (from Step 5) and solved for $\pi$:

$$\frac{\sqrt{3}}{8} + \frac{\pi}{12} = \text{Series Sum}$$ $$\frac{\pi}{12} = \text{Series Sum} - \frac{\sqrt{3}}{8}$$ $$\pi = 12 \left( \text{Series Sum} - \frac{\sqrt{3}}{8} \right)$$ $$\pi = 12 \left[ \left( \frac{1}{2} - \frac{1}{48} - \frac{1}{1280} - \dots \right) - \frac{\sqrt{3}}{8} \right]$$

Using just the first few terms of the series, Newton achieved an approximation of $\pi \approx 3.14161$.