The value of \(\sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots } } }\)
Lies between 1 and 2
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:
| Option | Statement | Evaluation (with \(x \approx 1.618\)) | Conclusion |
|---|---|---|---|
| 1 | Equals to 1 | \(1.618 = 1\) | False |
| 2 | Lies between 0 and 1 | \(0 < 1.618 < 1\) | False |
| 3 | Lies between 1 and 2 | \(1 < 1.618 < 2\) | True |
| 4 | Is greater than 2 | \(1.618 > 2\) | False |
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
| Concept | Description | How it Applies Here |
|---|---|---|
| Infinite Nested Radical | An expression where a square root is nested inside itself infinitely. | The problem is of this form: \(\sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}\). |
| Self-Similarity | The structure of the expression repeats itself. | The part under the outermost root is the same as the whole expression. |
| Setting up an Equation | Assigning a variable (e.g., \(x\)) to the expression and using self-similarity to write an equation. | We set \(x = \sqrt{1 + x}\). |
| Solving Quadratic Equation | Using methods like the quadratic formula to find the roots of \(ax^2 + bx + c = 0\). | The equation \(x^2 - x - 1 = 0\) is solved using the quadratic formula. |
| Validating Solutions | Checking if the solutions obtained satisfy the conditions of the original problem (e.g., square roots give non-negative values). | We rejected the negative solution as the square root expression must be non-negative. |
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.
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