[Math] Determine convergents for square root

Question

All square roots can be represented as a continued fraction.
The fraction can be calculated to $n$ terms (e.g. $\sqrt{2}$ is $[1; 2, 2, 2, 2…]$)

So the continued fraction for $\sqrt{2}$ to $3$ terms would be
$$1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + 1}}} = \frac{17}{12}.$$

Additionally, the next convergent for square root of $2$ can be calculated with the formula $$\frac{numerator + 2 * denominator}{numerator + denominator}$$
How would I find the simplified continued fraction to $Y$ terms for the square root of $N$? So $f(2, 3)$ would be $\frac{17}{12}$.

If a similar formula to calculating the next convergent of $\sqrt{2}$ exists for all square roots, please include that formula.

Answer 1

This tutorial site on continued fractions presents a generalized means of calculating the successive convergents for a continued fraction from left to right, each in terms of the numerator and denominators of the previous convergent:

If our continued fraction is [a₀; a₁, a₂, ..., ] then the successive convergents are available (defining N_-1 ≡ 1 and D_-1 ≡ 0) as:

0. C₀ ≡ N₀ / D₀ = a₀ / 1 = a₀;

1. C₁ ≡ N₁ / D₁ = (a₁⋅N₀ + N_-1) / (a₁⋅D₀ + D_-1) = (a₁⋅a₀ + 1) / (a₁⋅c + 0)

    :

    n. C_n ≡ N_n / D_n = (a_n⋅N_n-1 + N_n-2) / (a_n⋅D_n-1 + D_n-2)

For your example of $$\sqrt{2} = [1; (2)]$$ one obtains in succession:

0. a₀=1; C₀ ≡ 1 / 1 = 1;

1. a₁=2; C₁ ≡ (2.1 + 1) / (2.1 + 0) = 3 / 2;

2. a₂=2; C₂ ≡ (2.3 + 1) / (2.2 + 1) = 7 / 5;

3. a₃=2; C₃ ≡ (2.7 + 3) / (2.5 + 2) = 17 / 12;

4. a₄=2; C₄ ≡ (2.17 + 7) / (2.12 + 5) = 41 / 29.

Likewise for $$\sqrt{15} = [3; (1, 6)]$$

0. a₀=3; C₀ ≡ 3 / 1 = 3;

1. a₁=1; C₁ ≡ (1.3 + 1) / (1.1 + 0) = 4 / 1;

2. a₂=6; C₂ ≡ (6.4 + 3) / (6.1 + 1) = 27 / 7;

3. a₃=1; C₃ ≡ (1.27 + 4) / (1.7 + 1) = 31 / 8;

4. a₄=6; C₄ ≡ (6.31 + 27) / (6.8 + 7) = 213 / 55.