Square Root of Two by Linear Approximation with Calculus
We can use the derivative from calculus to calculate the square root two. We need to take the derivative of the function \(f(x)=\sqrt{x}\), use the equation of the tangent line, and use values that are near two that we can actually find the square root of without special techniques.
The equation of the tangent line at \(x=a\) is normally written like
$$y-f(a)=f'(a)(x-a).$$For our purposes, we will use the linearization function normally presented in calculus textbooks (it's the same!),
$$L(x)=f(a)+f'(a)(x-a).$$The derivative of \(f(x)=\sqrt{x}\) is
$$f'(x)=\frac{1}{2\sqrt{x}}.$$Our linearization function is now
$$L(x)=\sqrt{a} + \frac{1}{2\sqrt{a}} \left( x - a \right).$$The fun part now is coming up with numbers close to two that we can find the square root of. Let's make a list of perfect squares, so we can stare at it.
$$1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,\ldots$$One is pretty close to two, but we can do so much better. Let's try taking ratios that look like they might be close to two.
$$\frac{9}{4} = 2.25$$ $$\frac{49}{25} = 1.96$$ $$\frac{100}{49} \approx 2.041$$Let's choose \(a=\frac{49}{25}\) because that seems to be closest number to two that I see after some examination. You can always go further out to try and find more. You might also notice there are some repeats.
Feeding this into our linearization function, we get
$$L(x)=\sqrt{\frac{49}{25}} + \frac{1}{2\sqrt{\frac{49}{25}}} \left( x - \frac{49}{25} \right).$$After simplifying,
$$L(x)=\frac{7}{5} + \frac{5}{14} \left( x - \frac{49}{25} \right).$$Now let \(x=2\).
$$L(2)=\frac{7}{5} + \frac{5}{14} \left( \frac{50}{25} - \frac{49}{25} \right)$$ $$ =\frac{7}{5} + \frac{5}{14} \left( \frac{1}{25} \right)$$ $$ =\frac{7}{5} + \frac{1}{70}$$ $$ =\frac{98}{70} + \frac{1}{70}$$ $$L(2)=\frac{99}{70} = 1.4142857142857\ldots = 1.4\overline{142857}$$The actual square root of two is
$$\sqrt{2} = 1.414213562373095\ldots$$