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Let \(c=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}\). How is \(c^{2}\) related to \(c\) ? What is the value of \(c\) ?

Short Answer

Expert verified
The number \(c = \frac{1 + \sqrt{5}}{2} \approx 1.618\) and \(c^{2}\) is related to \(c\) by the equation \(c^{2} = 1 + c\).

Step by step solution

01

Identify the recursive pattern

Here, the given sequence is \(c = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}\). The sequence under the square root sign after '1+' can be represented as 'c' itself, given it is a recursive pattern.
02

Formulate an equation

We represent the entire sequence after 'c =' as 'c' itself, yielding the equation \(c = \sqrt{1+c}\)
03

Simplify the Equation

Square both sides of the equation to get rid of the square root yielding the equation \(c^2 = 1 + c\).
04

Solve for 'c'

Rearrange the equation from Step 3 to \(c^2 - c - 1 = 0\). Solving this quadratic equation yields two results. Using the quadratic formula \((-b \pm \sqrt{b^2 - 4ac}) / (2a)\), we get two solutions \(c = \frac{1 \pm \sqrt{5}}{2}\). However, since \(c\) is positive, its valid value is \(c = \frac{1 + \sqrt{5}}{2}\), which is approximately 1.618.
05

Answer the original question

For the second part of the question, \(c^{2}\) is related to \(c\) by the equation \(c^{2} = 1 + c\), as derived in step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are a type of polynomial equation where the highest power of the variable is squared. A general form of a quadratic equation can be expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable. In the context of our exercise, we had to rearrange \( c^2 = 1 + c \) into the standard quadratic form to solve for \( c \).

To derive the solutions to the quadratic equation \( c^2 - c - 1 = 0 \), we use the quadratic formula: \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Key points to remember when dealing with quadratic equations:
  • Identify the coefficients \( a \), \( b \), and \( c \).
  • Apply the quadratic formula or factorize to find the values of \( x \).
  • Consider the quadratic discriminant \( b^2 - 4ac \) to determine the nature of the roots (real and distinct, real and equal, or complex).
In our case, the discriminant is positive, which indicates we have two distinct real solutions. We select the positive solution based on the context of the problem.
Radicals in Equations
Radicals, often seen as the square root or nth root in equations, indicate operations that reverse the process of exponentiation. In this exercise, radical expressions are nested within each other, creating a complex recursive sequence.

To deal with radicals in equations effectively:
  • Simplify the expression, if possible, by isolating the radical on one side.
  • Squaring both sides often helps eliminate the radical, but be careful as it might introduce extraneous solutions.
  • Check your solutions by substituting back into the original equation to ensure they are valid.
When given \( c = \sqrt{1 + c} \), the radical sign presents a recursive pattern. It implies that the sequence continues infinitely inward. By treating the entire radical sequence as \( c \) itself, we were able to simplify and eliminate the radical to solve further using algebraic methods.
Fixed Points in Sequences
A fixed point in sequences is an element of the sequence that remains constant throughout the iterations. In recursive sequences, such as the one in our problem, identifying the fixed point helps determine the behavior and eventual steady state of the sequence.

In this exercise, the expression \( c = \sqrt{1 + c} \) converges to a fixed point, where implementing values of \( c \) back into the equation yield the same \( c \).

To understand fixed points in sequences:
  • Analyze the recursive pattern to determine if the sequence tends toward a limit.
  • Equate all elements under consideration to solve for the fixed point value.
  • Understand that only positive, real values are typically meaningful in convergence problems.
Thanks to the recursive nature of the sequence and understanding fixed points, we deduced that \( c \) converges to \( \frac{1 + \sqrt{5}}{2} \), showcasing the power and application of fixed points in mathematics.

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Most popular questions from this chapter

For \(n \geq 0\), let us toss a coin \(2 n\) times. a) If \(a_{n}\) is the number of sequences of \(2 n\) tosses where \(n\) heads and \(n\) tails occur, find \(a_{n}\) in terms of \(n\). b) Find constants \(r, s\), and \(t\) so that \((r+s x)^{t}=f(x)=\) \(\sum_{n=0}^{\infty} a_{n} x^{n}\) c) Let \(b_{n}\) denote the number of sequences of \(2 n\) tosses where the numbers of heads and tails are equal for the first time only after all \(2 n\) tosses have been made. (For example, if \(n=3\), then HHHTTT and HHTHTT are counted in \(b_{n}\). but HTHHTT and HHTTHT are not.) Define \(b_{0}=0\) and show that for all \(n \geq 1\), $$ a_{n}=a_{0} b_{n}+a_{1} b_{n-1}+\cdots+a_{n-1} b_{1}+a_{n} b_{0} $$ d) Let \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\). Show that \(g(x)=1-1 / f(x)\), and then solve for \(b_{n}, n \geq 1\).

a) For \(n \in \mathbf{Z}^{+}\), let \(a_{n}\) count the number of palindromes of \(2 n\). Then \(a_{n+1}=2 a_{n}, n \geq 1, a_{1}=2\). Solve this first-order recurrence relation for \(a_{n}\). b) For \(n \in \mathbf{Z}^{+}\), let \(b_{n}\) count the number of palindromes of \(2 n-1\). Set up and solve a first-order recurrence relation for \(b_{n}\).

Verify that for all \(n \geq 0\), $$ \frac{1}{2}\left(\frac{1}{2 n+1}\right)\left(\begin{array}{c} 2 n+2 \\ n+1 \end{array}\right)=\left(\frac{1}{n+1}\right)\left(\begin{array}{c} 2 n \\ n \end{array}\right) $$

Solve the recurrence relation \(a_{n+2}-6 a_{n+1}+9 a_{n}=\) \(3\left(2^{n}\right)+7\left(3^{n}\right)\), where \(n \geq 0\) and \(a_{0}=1, a_{1}=4\)

Let \(a, b\) be fixed nonzero real numbers. Determine \(x_{n}\) if \(x_{n}=x_{n-1} x_{n-2}, n \geq 2, x_{0}=a, x_{1}=b .\)

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