Understanding the Infinite Nested Square Root Problem

The problem asks for the value of an expression with an infinite sequence of nested square roots: \(\sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } }\). This type of problem can be solved by recognizing the self-similar nature of the expression.

Solving the Infinite Radical Expression

Let the value of the given expression be \(x\). So, we have:

\[x = \sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } }\]

Since the expression under the outermost square root sign is exactly the same as the original expression \(x\), we can rewrite the equation as:

\[x = \sqrt{1 + x}\]

To eliminate the square root, we square both sides of the equation:

\[x^2 = (\sqrt{1 + x})^2\]

\[x^2 = 1 + x\]

Now, we rearrange the terms to form a quadratic equation:

\[x^2 - x - 1 = 0\]

Finding the Value Using the Quadratic Formula

We can solve this quadratic equation of the form \(ax^2 + bx + c = 0\) using the quadratic formula, which is:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In our equation, \(x^2 - x - 1 = 0\), we have \(a=1\), \(b=-1\), and \(c=-1\). Substituting these values into the formula:

\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}\]

\[x = \frac{1 \pm \sqrt{1 + 4}}{2}\]

\[x = \frac{1 \pm \sqrt{5}}{2}\]

This gives us two potential solutions for \(x\):

• \(x_1 = \frac{1 + \sqrt{5}}{2}\)
• \(x_2 = \frac{1 - \sqrt{5}}{2}\)

Determining the Valid Solution

The original expression \(\sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } }\) involves taking square roots of terms that are always positive (\(1 + \text{a positive value or zero}\)). The square root symbol (\(\sqrt{\phantom{x}}\)) typically denotes the principal (non-negative) square root. Therefore, the value of \(x\) must be non-negative.

Let's look at the two potential solutions:

• \(x_1 = \frac{1 + \sqrt{5}}{2}\). Since \(\sqrt{5}\) is a positive number (approximately 2.236), \(1 + \sqrt{5}\) is positive, and \(\frac{1 + \sqrt{5}}{2}\) is positive.
• \(x_2 = \frac{1 - \sqrt{5}}{2}\). Since \(\sqrt{5} \approx 2.236\), \(1 - \sqrt{5} \approx 1 - 2.236 = -1.236\). Therefore, \(\frac{1 - \sqrt{5}}{2}\) is a negative number.

Since the value of the nested square root expression must be non-negative, we reject the negative solution \(x_2\). The valid value is \(x = \frac{1 + \sqrt{5}}{2}\).

Analyzing the Value and Options

The value \(x = \frac{1 + \sqrt{5}}{2}\) is the golden ratio, often denoted by \(\phi\).

To determine which option is correct, let's approximate the value:

\(\sqrt{5} \approx 2.236\)

\(x \approx \frac{1 + 2.236}{2} = \frac{3.236}{2} \approx 1.618\)

Now, let's compare this approximate value with the given options:

The value \(x \approx 1.618\) clearly lies between 1 and 2.

Conclusion on the Infinite Radical Value

The value of \(\sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } }\) is \(\frac{1 + \sqrt{5}}{2}\), which is approximately 1.618. This value lies between 1 and 2.

Revision Table: Infinite Nested Radicals

Additional Information on Nested Roots and Golden Ratio

The value \(\frac{1 + \sqrt{5}}{2}\) is known as the golden ratio (\(\phi\)). It appears in various areas of mathematics, nature, and art. Problems involving infinite nested radicals often converge to a specific value.

Consider a more general form: \(y = \sqrt{a + \sqrt{a + \sqrt{a + \cdots}}}\). Setting \(y = \sqrt{a+y}\), we get \(y^2 = a+y\), or \(y^2 - y - a = 0\). Using the quadratic formula, \(y = \frac{1 \pm \sqrt{1 + 4a}}{2}\). Since \(y\) must be positive, the solution is \(y = \frac{1 + \sqrt{1 + 4a}}{2}\). In our specific problem, \(a=1\), which gives \(x = \frac{1 + \sqrt{1 + 4(1)}}{2} = \frac{1 + \sqrt{5}}{2}\).

Another type of infinite radical is \(z = \sqrt{a \cdot \sqrt{a \cdot \sqrt{a \cdot \cdots}}}\). Squaring gives \(z^2 = a \cdot \sqrt{a \cdot \sqrt{a \cdot \cdots}} = a \cdot z\). If \(z \neq 0\), we can divide by \(z\) to get \(z=a\). For example, \(\sqrt{2 \cdot \sqrt{2 \cdot \sqrt{2 \cdot \cdots}}} = 2\).

These infinite expressions are examples of how sequences defined by recurrence relations can converge to a limit, which is found by solving an equation derived from the self-similarity.