Question

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0},\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G.)\) a. Show that \(a_{n} > b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

Short answer

Question: Show that the sequences \({a_n}\) and \({b_n}\) described above have the properties stated in the problem. Answer: We can show that the sequences \({a_n}\) and \({b_n}\) have the desired properties by following the steps outlined in the solution. Firstly, we can prove by induction that \(a_n > b_n\) for all \(n\). Next, we can show that \({a_n}\) is a decreasing sequence while \({b_n}\) is an increasing sequence, also using induction. Since these sequences are bounded, they converge to limits \(A\) and \(B\). We can find an expression for the difference \(a_{n+1}-b_{n+1}\) and show that their limits are equal. Finally, we can estimate AGM(12, 20) and Gauss' constant.

Step-by-step solution

Show that \(a_n > b_n\) for all \(n\)

We are given that \(a_0>b_0>0\). Let's prove \(a_n > b_n\) for all \(n\) by induction. Base case: n=0 Since we are given that \(a_0>b_0\), the base case holds. Inductive step: Assume \(a_n > b_n\) for some \(n\). Now let's prove that \(a_{n+1} > b_{n+1}\). $$ a_{n+1} = \frac{a_n + b_n}{2} > \frac{b_n + b_n}{2} = b_n $$ $$ b_{n+1} = \sqrt{a_n b_n} < \sqrt{a_n a_n} = a_n $$ Thus, we have shown that if \(a_n > b_n\), then \(a_{n+1} > b_{n+1}\), and by the induction principle, \(a_n > b_n\) for all \(n\).

Show that \({a_n}\) is decreasing and \({b_n}\) is increasing

We will again use induction. Base case: n=0 \( a_1 = \frac{a_0 + b_0}{2} < a_0\) since \(a_0>b_0>0\) \( b_1 = \sqrt{a_0 b_0} > b_0\) since \(a_0 > b_0>0\) Inductive step: Assume \({a_{n+1}} < {a_n}\) and \({b_{n+1}} > {b_n}\) for some \(n\). Now let's prove that \({a_{n+2}} < {a_{n+1}}\) and \({b_{n+2}} > {b_{n+1}}\). $$ a_{n+2}=\frac{a_{n+1}+b_{n+1}}{2} < \frac{a_{n}+a_{n}}{2} = a_{n} $$ $$ b_{n+2}=\sqrt{a_{n+1} b_{n+1}} > \sqrt{b_{n+1} b_{n+1}} = b_{n+1} $$ Thus, we have shown that if \({a_n}\) is decreasing and \({b_n}\) is increasing for some \(n\), they remain decreasing and increasing for \(n+1\). By induction, \({a_n}\) is a decreasing sequence and \({b_n}\) is an increasing sequence for all \(n\).

Conclude that \({a_n}\) and \({b_n}\) converge

Since \({a_n}\) is a decreasing sequence and bounded below by \({b_n}\) (as we proved in Step 1), it converges to some limit \(A\). Similarly, since \({b_n}\) is an increasing sequence and bounded above by \({a_n}\), it converges to some limit \(B\).

Show that \(a_{n+1}-b_{n+1}

Now, let's find an expression for \(a_{n+1}-b_{n+1}\): $$ a_{n+1}-b_{n+1}=\frac{a_n+b_n}{2}-\sqrt{a_nb_n}=\frac{a_n-b_n}{2}-\frac{a_n-b_n}{2}\left(1-\frac{2\sqrt{a_nb_n}}{a_n+b_n}\right) $$ Since \(a_n > b_n > 0\), we have \(1-\frac{2\sqrt{a_nb_n}}{a_n+b_n}<\frac{1}{2}\). So, \(a_{n+1}-b_{n+1}<\left(a_n-b_n\right) / 2\). As \(n \to \infty\), \(a_n - b_n \to 0\). Therefore, \(\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} b_n\).

Estimate AGM(12, 20) and estimate Gauss' constant

To estimate AGM(12, 20), we'll start with the given values \(a_0 = 12\) and \(b_0 = 20\). Let's calculate a few terms of the sequences \({a_n}\) and \({b_n}\): \(a_1 = \frac{12 + 20}{2} = 16\) \(b_1 = \sqrt{12 \cdot 20} \approx 15.652\) \(a_2 = \frac{16 + 15.652}{2} \approx 15.826\) \(b_2 = \sqrt{16 \cdot 15.652} \approx 15.826\) \(a_3 = \frac{15.826 + 15.826}{2} = 15.826\) \(b_3 = \sqrt{15.826 \cdot 15.826} = 15.826\) We can see that the two sequences have already converged in just a few iterations. So, AGM(12, 20) = 15.826. To estimate Gauss' constant, which is \(1/AGM(1, \sqrt{2})\), start with \(a_0 = 1\) and \(b_0 = \sqrt{2}\): \(a_1 = \frac{1 + \sqrt{2}}{2} \approx 1.207\) \(b_1 = \sqrt{1 \cdot \sqrt{2}} \approx 1.189\) \(a_2 = \frac{1.207 + 1.189}{2} \approx 1.198\) \(b_2 = \sqrt{1.207 \cdot 1.189} \approx 1.198\) \(a_3 = \frac{1.198 + 1.198}{2} = 1.198\) \(b_3 = \sqrt{1.198 \cdot 1.198} = 1.198\) As the sequences converge, AGM(1, \(\sqrt{2}\)) = 1.198. Therefore, Gauss' constant is approximately \(1/1.198 \approx 0.835\).

Key concepts

Sequences and Series

In mathematics, sequences and series are essential foundations in understanding change and patterns. A sequence is an ordered list of numbers, casually known as a list of terms, which can be finite or infinite. When you see sequences like \(a_n\) and \(b_n\) in exercises, they represent continuous flows or progressions of numbers defined by a specific mathematical rule.
For the given problem, we define two sequences whereby \(a_{n+1} = \frac{a_n + b_n}{2}\) and \(b_{n+1} = \sqrt{a_n b_n}\). These equations guide how each term relates to its predecessor, setting up a systematic way to understand how each progresses.
A series, on the other hand, is akin to summing up the terms of a sequence, but in this problem, we're focused on the behavior of the sequences themselves, particularly examining their paths and limits as terms proceed.

Convergence

Convergence is a core concept in calculus and analysis, describing how a sequence approaches a specific value as it progresses.
In the context of our problem, we established that the sequence \(a_n\) is decreasing while \(b_n\) is increasing. Intuitively, this bound means they must converge towards each other. These properties make the lines of \(a_n\) and \(b_n\) meet—or converge—at a particular point or value.
Because \(a_n\) is bounded below by \(b_n\) and \(b_n\) is bounded above by \(a_n\), they naturally find a common limit. As \(n\rightarrow\infty\), the difference between the sequences diminishes, leading both to ultimately meet at the same limit. This shared limit is known as the Arithmetic-Geometric Mean (AGM).
Convergence not only assures us about stability regarding the terms, but it allows us to derive practical values out of potentially infinite processes.

Arithmetic Mean

The arithmetic mean is a fundamental concept in mathematics that provides a simple calculation to find an average value in a set of numbers. It is computed by adding together all the numbers in the set and dividing by the count of the numbers.
In our sequence, the formula for updating \(a_n\) is based on the arithmetic mean: \(a_{n+1} = \frac{a_n + b_n}{2}\). This indicates that each successive term \(a_{n+1}\) is the arithmetic mean of the current terms \(a_n\) and \(b_n\).
The principle behind this is equal distribution, as it pulls every term towards the center or average, thereby potentially smoothing out any fluctuations by averaging. In our problem's context, the arithmetic mean aids in moving the sequence towards a convergence point by balancing the overall values across terms.

Geometric Mean

The geometric mean provides another perspective on averaging by multiplying n numbers and then taking the nth root, which is suited for sets of positive numbers. It is particularly useful in various fields like finance and population studies, where the relative change matters more than absolute change.
For our sequences in the problem, \(b_n\) is partly calculated using the geometric mean: \(b_{n+1} = \sqrt{a_n b_n}\). This step involves multiplying two values (harvested from the sequences) and finding their square root, thus inherently scaling quantities down towards a common value without substantial variance influence.
The geometric mean helps ensure the values grow proportionally. By harmonizing the terms' relative sizes, it aligns closely with the trends represented by the arithmetic mean, finally assisting both sequences to converge. Through these interactions, the arithmetic-geometric disparity is minimized, pointing each sequence towards the AGM.