The value of the infinite product \({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \) is
36
Calculating the Value of an Infinite Product
The problem asks for the value of an infinite product involving the base 6 raised to a sequence of exponents.
The given infinite product is:
\({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \)
Using the property of exponents, \(a^m \times a^n = a^{m+n}\), an infinite product of the form \(\prod_{n=1}^{\infty} a^{x_n}\) can be written as \(a^{\sum_{n=1}^{\infty} x_n}\). In this case, the base is \(a=6\), and the exponents are \(x_n\). So, the value of the product is \(6^S\), where \(S\) is the sum of the infinite series of exponents.
The sequence of exponents is:
- \(a_1 = \frac{1}{2}\)
- \(a_2 = \frac{1}{2}\)
- \(a_3 = \frac{3}{8}\)
- \(a_4 = \frac{1}{4}\)
- and so on...
We need to find the sum \(S = \frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \ldots\).
Finding the Pattern in the Exponents Sequence
Let's examine the given terms in the sequence of exponents to find a pattern. Let the sequence be denoted by \(a_n\).
\(a_1 = \frac{1}{2}\)
\(a_2 = \frac{1}{2}\)
\(a_3 = \frac{3}{8}\)
\(a_4 = \frac{1}{4}\)
Let's try to find if this sequence follows a linear recurrence relation of the form \(a_n = c_1 a_{n-1} + c_2 a_{n-2}\) for \(n \ge 3\). We can use the given terms to set up a system of equations for the coefficients \(c_1\) and \(c_2\).
- Using \(a_1, a_2, a_3\): For \(n=3\), \(a_3 = c_1 a_2 + c_2 a_1\). Substituting the values, we get:
\(\frac{3}{8} = c_1 \left(\frac{1}{2}\right) + c_2 \left(\frac{1}{2}\right)\)
Multiplying the equation by 8 to clear the denominators:
\(3 = 4c_1 + 4c_2\)
Dividing by 4:
\(c_1 + c_2 = \frac{3}{4}\) (Equation 1) - Using \(a_2, a_3, a_4\): For \(n=4\), \(a_4 = c_1 a_3 + c_2 a_2\). Substituting the values, we get:
\(\frac{1}{4} = c_1 \left(\frac{3}{8}\right) + c_2 \left(\frac{1}{2}\right)\)
Multiplying the equation by 8 to clear the denominators:
\(2 = 3c_1 + 4c_2\) (Equation 2)
Solving for the Recurrence Coefficients
Now we solve the system of linear equations for \(c_1\) and \(c_2\):
\(c_1 + c_2 = \frac{3}{4}\)
\(3c_1 + 4c_2 = 2\)
From Equation 1, we can express \(c_1\) as \(c_1 = \frac{3}{4} - c_2\). Substitute this into Equation 2:
\(3\left(\frac{3}{4} - c_2\right) + 4c_2 = 2\)
\(\frac{9}{4} - 3c_2 + 4c_2 = 2\)
\(c_2 = 2 - \frac{9}{4} = \frac{8}{4} - \frac{9}{4} = -\frac{1}{4}\)
Now substitute the value of \(c_2\) back into the expression for \(c_1\):
\(c_1 = \frac{3}{4} - c_2 = \frac{3}{4} - \left(-\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1\)
So, the recurrence relation for the exponents is \(a_n = 1 \cdot a_{n-1} - \frac{1}{4} a_{n-2}\), which simplifies to \(a_n = a_{n-1} - \frac{1}{4} a_{n-2}\) for \(n \ge 3\). The initial conditions are \(a_1 = \frac{1}{2}\) and \(a_2 = \frac{1}{2}\).
Summing the Infinite Series of Exponents
We need to find the sum \(S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \ldots\). The terms \(a_n\) are generated by the recurrence \(a_n - a_{n-1} + \frac{1}{4} a_{n-2} = 0\) for \(n \ge 3\).
Assuming the series converges, we can sum the recurrence relation from \(n=3\) to infinity:
\(\sum_{n=3}^{\infty} \left(a_n - a_{n-1} + \frac{1}{4} a_{n-2}\right) = \sum_{n=3}^{\infty} 0 = 0\)
This can be split into individual sums:
\(\sum_{n=3}^{\infty} a_n - \sum_{n=3}^{\infty} a_{n-1} + \frac{1}{4} \sum_{n=3}^{\infty} a_{n-2} = 0\)
Let \(S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \ldots\). We can express the sums in terms of \(S\) and the initial terms:
- \(\sum_{n=3}^{\infty} a_n = a_3 + a_4 + a_5 + \ldots = S - a_1 - a_2\)
- \(\sum_{n=3}^{\infty} a_{n-1} = a_2 + a_3 + a_4 + \ldots = S - a_1\)
- \(\sum_{n=3}^{\infty} a_{n-2} = a_1 + a_2 + a_3 + \ldots = S\)
Substitute these expressions back into the summed recurrence relation:
\((S - a_1 - a_2) - (S - a_1) + \frac{1}{4} S = 0\)
\(S - a_1 - a_2 - S + a_1 + \frac{1}{4} S = 0\)
The \(S\) terms cancel out, and the \(a_1\) terms cancel out:
\(-a_2 + \frac{1}{4} S = 0\)
Solving for \(S\):
\(\frac{1}{4} S = a_2\)
\(S = 4 a_2\)
Using the value of the second term, \(a_2 = \frac{1}{2}\), we find the sum of the exponents:
\(S = 4 \times \frac{1}{2} = 2\)
To confirm the series converges, we look at the characteristic equation of the recurrence: \(r^2 - c_1 r - c_2 = 0\), which is \(r^2 - 1 r - (-\frac{1}{4}) = 0\), or \(r^2 - r + \frac{1}{4} = 0\). This equation factors as \(\left(r - \frac{1}{2}\right)^2 = 0\). The roots are \(r = \frac{1}{2}\) with multiplicity 2. Since the modulus of the root \(|\frac{1}{2}| = \frac{1}{2}\) is less than 1, the terms \(a_n\) approach 0 as \(n \to \infty\), and the sum \(S\) converges to a finite value.
Calculating the Infinite Product Value
The value of the infinite product is \(6^S\), where \(S=2\).
\(6^S = 6^2 = 36\)
Final Answer
The value of the infinite product \({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \) is 36.
Revision Table: Infinite Product Calculation
| Concept | Explanation |
|---|---|
| Infinite Product Property | An infinite product of the form \(\prod_{n=1}^\infty a^{x_n}\) is equal to \(a^{\sum_{n=1}^\infty x_n}\), provided the sum of exponents converges. |
| Sum of Exponents Series | The problem reduces to finding the sum of the infinite series of exponents: \(\frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \ldots\). |
| Linear Recurrence Relation | The sequence of exponents \((a_n)\) follows \(a_n = a_{n-1} - \frac{1}{4} a_{n-2}\) for \(n \ge 3\), with \(a_1 = \frac{1}{2}, a_2 = \frac{1}{2}\). |
| Summing Recurrence Series | For a convergent series from a homogeneous linear recurrence, summing the recurrence equation term by term can determine the total sum \(S\). In this case, \(-a_2 + \frac{1}{4} S = 0\). |
| Calculated Sum of Exponents | From \(-a_2 + \frac{1}{4} S = 0\) and \(a_2 = \frac{1}{2}\), we get \(S = 4a_2 = 4 \times \frac{1}{2} = 2\). |
| Final Product Value | The infinite product is \(6^S = 6^2 = 36\). |
Additional Information: Infinite Series and Recurrences
Infinite products are extensions of finite products. An infinite product \(\prod_{n=1}^{\infty} c_n\) converges to a non-zero value if and only if the series \(\sum_{n=1}^{\infty} (c_n - 1)\) converges. For positive terms \(c_n\), it converges if and only if \(\sum_{n=1}^{\infty} \ln(c_n)\) converges. In our case, \(c_n = 6^{a_n}\), so \(\ln(c_n) = a_n \ln(6)\). The sum \(\sum a_n \ln(6) = \ln(6) \sum a_n\). Since \(\ln(6)\) is a constant, the convergence of the infinite product depends directly on the convergence of the series of exponents \(\sum a_n\).
Linear recurrence relations describe sequences where each term is a linear combination of previous terms. The behaviour of these sequences is largely determined by the roots of their characteristic equation. For a second-order recurrence \(a_n = c_1 a_{n-1} + c_2 a_{n-2}\), the characteristic equation is \(r^2 - c_1 r - c_2 = 0\). If all roots have a magnitude less than 1, the terms \(a_n\) approach 0 as \(n\) goes to infinity, which is a necessary condition for the series \(\sum a_n\) to converge.
The method used to find the sum \(S\) by summing the recurrence relation is a valid technique when the series converges. It effectively collapses the infinite series into a simple equation involving the sum \(S\) and the initial terms of the sequence.
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