Problem 887: Chained Radicals

Mathematical Analysis

Theorem 1 (Fixed Point Convergence)

Proof. If the nested radical converges to $x$, then $x = \sqrt{a + x}$, so $x^2 = a + x$, giving $x^2 - x - a = 0$. By the quadratic formula: $x = \frac{1 \pm \sqrt{1 + 4a}}{2}$. Since $x \geq 0$, we take the positive root. $\square$

Theorem 2 (Convergence Guarantee)

For $a \geq 0$, define $x_0 = 0$ and $x_{n+1} = \sqrt{a + x_n}$. Then $\{x_n\}$ converges monotonically to $R(a)$.

Proof. The function $g(x) = \sqrt{a + x}$ is a contraction on $[0, R(a)]$ since $|g'(x)| = \frac{1}{2\sqrt{a+x}} < 1$ for $x \geq 0$ and $a > 0$. By the Banach fixed point theorem, iteration converges. Monotonicity: $x_1 = \sqrt{a} > 0 = x_0$, and $g$ is increasing, so $x_2 > x_1$, etc. $\square$

Theorem 3 (Rate of Convergence)

since $|g'(R(a))| = \frac{1}{2R(a)} < 1$ (linear convergence).

Theorem 4 (Generalized Nested Radical)

For the Ramanujan radical with varying terms:

More generally, $\sqrt{a_1 + \sqrt{a_2 + \sqrt{a_3 + \cdots}}}$ converges when $a_n = O(n^2)$.

Lemma (Integer Solutions)

$R(a)$ is an integer $n$ iff $n^2 - n = a$, i.e., $a = n(n-1)$ for $n \geq 1$.

Concrete Numerical Examples

Convergence Example ($a = 2$)

| $n$ | $x_n = \sqrt{2 + x_{n-1}}$ | Error $|x_n - 2|$ | |:-:|:-:|:-:| | 0 | 0 | 2.000 | | 1 | 1.414 | 0.586 | | 2 | 1.848 | 0.152 | | 3 | 1.961 | 0.039 | | 4 | 1.990 | 0.010 | | 5 | 1.997 | 0.003 |

Golden Ratio Connection

For $a = 1$: $R(1) = \frac{1 + \sqrt{5}}{2} = \phi \approx 1.61803$, the golden ratio. This means:

Generalization: Different Terms

Varying nested radicals

converges when $a_n \leq C \cdot n^2$ for some constant $C$ (Herschfeld’s theorem).

Ramanujan’s Famous Identity

Proof. Define $f(n) = \sqrt{1 + n\sqrt{1 + (n+1)\sqrt{\cdots}}}$. One can show $f(n) = n+1$ by verifying $(n+1)^2 = 1 + n \cdot f(n+1) = 1 + n(n+2) = n^2 + 2n + 1$. $\square$

Cube Root Variant

The equation $x = \sqrt[3]{a + x}$ gives $x^3 - x - a = 0$. By Cardano’s formula, the real root can be expressed in terms of $a$.

Continued Fraction Connection

For $a = 1$: $\phi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}$ is the continued fraction, while $\phi = \sqrt{1 + \sqrt{1 + \cdots}}$ is the nested radical. Both converge to the golden ratio, connecting two fundamental iterative structures.

Complexity Analysis

Answer

/*
 * Problem 887: Chained Radicals
 * R(a) = sqrt(a + sqrt(a + ...)) = (1 + sqrt(1 + 4a)) / 2.
 */
#include <bits/stdc++.h>
using namespace std;

double nested_radical(double a) {
    return (1.0 + sqrt(1.0 + 4.0 * a)) / 2.0;
}

double nested_iterative(double a, int iters = 100) {
    double x = 0;
    for (int i = 0; i < iters; i++) x = sqrt(a + x);
    return x;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cout << fixed << setprecision(10);
    cout << "=== Nested Radical ===" << endl;
    for (double a : {0.0, 1.0, 2.0, 3.0, 6.0, 12.0, 20.0, 100.0}) {
        double exact = nested_radical(a);
        double iter = nested_iterative(a, 200);
        cout << "a=" << a << ": exact=" << exact << " iter=" << iter
             << " diff=" << scientific << abs(exact - iter) << fixed << endl;
    }

    cout << "\n=== Golden Ratio ===" << endl;
    cout << "R(1) = " << nested_radical(1.0) << endl;
    cout << "phi  = " << (1.0 + sqrt(5.0)) / 2.0 << endl;

    cout << "\n=== Integer Solutions ===" << endl;
    for (int n = 1; n <= 10; n++) {
        int a = n * (n - 1);
        cout << "R(" << a << ") = " << nested_radical(a) << " (n=" << n << ")" << endl;
    }

    cout << "\nAnswer: R(1) = phi = " << nested_radical(1.0) << endl;
    return 0;
}

"""
Problem 887: Chained Radicals
Nested radical: sqrt(a + sqrt(a + ...)) = (1 + sqrt(1+4a))/2.
"""

from math import sqrt

def nested_radical_exact(a):
    """Closed form: (1 + sqrt(1 + 4a)) / 2."""
    return (1 + sqrt(1 + 4 * a)) / 2

def nested_radical_iterative(a, iterations=100):
    """Fixed-point iteration: x_{n+1} = sqrt(a + x_n)."""
    x = 0.0
    for _ in range(iterations):
        x = sqrt(a + x)
    return x

def convergence_history(a, iterations=20):
    """Return list of iterates."""
    x = 0.0
    history = [x]
    for _ in range(iterations):
        x = sqrt(a + x)
        history.append(x)
    return history

def integer_solutions(max_n=20):
    """Find a where R(a) is integer: a = n(n-1)."""
    return [(n, n * (n - 1)) for n in range(1, max_n + 1)]

# --- Verification ---
print("=== Nested Radical Exact vs Iterative ===")
for a in [0, 1, 2, 3, 6, 12, 20, 30, 100]:
    exact = nested_radical_exact(a)
    iterative = nested_radical_iterative(a, 200)
    diff = abs(exact - iterative)
    print(f"  a={a:>3}: exact={exact:.6f}, iter={iterative:.6f}, diff={diff:.2e}")
    assert diff < 1e-10

print("\n=== Integer Solutions ===")
for n, a in integer_solutions(10):
    R = nested_radical_exact(a)
    print(f"  R({a}) = {R:.1f} (n={n})")
    assert abs(R - n) < 1e-10

print("\n=== Golden Ratio ===")
phi = nested_radical_exact(1)
phi_true = (1 + sqrt(5)) / 2
print(f"  R(1) = {phi:.10f}")
print(f"  phi  = {phi_true:.10f}")
assert abs(phi - phi_true) < 1e-12

print("\n=== Convergence for a=2 ===")
hist = convergence_history(2, 10)
for i, x in enumerate(hist):
    print(f"  x_{i} = {x:.6f}, error = {abs(x - 2):.6f}")

answer = nested_radical_exact(1)
print(f"\nAnswer: R(1) = phi = {answer:.10f}")

# --- 4-Panel Visualization ---