Solving the Wallis Product: Constructing Rectangles for $\pi/2$

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In summary, the problem concerns the Wallis Product \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots, where rectangles are constructed by attaching rectangles of area 1 alternately beside or on top of the previous rectangle. The goal is to find the limit of the ratios of width to height of these rectangles, which is ultimately shown to be \frac{\pi}{2}. By expressing each ratio in terms of heights and widths of previous terms and keeping the fractions factored,
  • #1
DivGradCurl
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Folks, I have a problem in my calculus textbook about the Wallis Product

[tex]\frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots[/tex]

"We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see figure). Find the limit of the ratios of width to height of these rectangles."

Here's what I've got:

[tex]R_1 = \frac{W_1}{H_1} = \frac{1}{1} = 1[/tex]

[tex]R_2 = \frac{W_2}{H_2} = \frac{W_1 + \frac{1}{H_2}}{H_1} = \frac{2}{1} = 2[/tex]

[tex]R_3 = \frac{W_3}{H_3} = \frac{W_2}{H_2 + \frac{1}{W_3}} = \frac{2}{3/2} = \frac{4}{3}[/tex]

[tex]R_4 = \frac{W_4}{H_4} = \frac{W_3 + \frac{1}{H_4}}{H_3} = \frac{8/3}{3/2} = \frac{16}{9}[/tex]

[tex]R_5 = \frac{W_5}{H_5} = \frac{W_4}{H_4 + \frac{1}{W_5}} = \frac{8/3}{15/8} = \frac{64}{45}[/tex]

Since all previous parts of this problem are about the Wallis Product, it would not be surprising to find out that the limit of the ratios of width to height of those rectangles is [tex]\frac{\pi}{2}[/tex]. It doesn't seem to be appropriate to simply state it, though.

If [tex]n[/tex] is even, then

[tex]W_n = W_{n-1} + \frac{1}{H_n}[/tex] and [tex]H_n = H_{n-1}[/tex]

and so

[tex]R_n = \frac{W_n}{H_n} = \frac{W_{n-1} + \frac{1}{H_n}}{H_{n-1}}[/tex]

If [tex]n[/tex] is odd, then

[tex]W_n = W_{n-1}[/tex] and [tex]H_n = H_{n-1} + \frac{1}{W_n}[/tex]

and so

[tex]R_n = \frac{W_n}{H_n} = \frac{W_{n-1}}{H_{n-1} + \frac{1}{W_n}}[/tex]

Thus, we find


[tex]\lim _{n\to \infty} \frac{W_{n-1} + \frac{1}{H_n}}{H_{n-1}} = \lim _{n\to \infty} \frac{W_{n-1}}{H_{n-1} + \frac{1}{W_n}}[/tex]

However, I don't know how to proceed so that I find

[tex]\lim _{n\to \infty} R_n = \frac{\pi}{2}[/tex]

Any help is highly appreciated. Thanks.

PS: THE FIGURE IS ATTACHED.
 

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  • #2
First, I think you want to express each ratio in terms of heights and widths of previous terms, not in terms of the current term. Then be sure to keep the H and W identifiable before reducing the fractions and leave the fractions factored. I added an intermediate step that you have done correctly, and idenitfied the W and H for each term in brackets.

[tex]R_1 = \frac{W_1}{H_1} = \frac{1}{1} = 1[/tex]

[tex]R_2 = \frac{W_2}{H_2} = \frac{W_1 + \frac{1}{H_1}}{H_1} = \frac{1 + 1}{1}= \left[\frac{2}{1}\right] = 1\cdot\frac{2}{1}[/tex]

[tex]R_3 = \frac{W_3}{H_3} = \frac{W_2}{H_2 + \frac{1}{W_2}} = \frac{2}{1 + 1/2} = \left[\frac{2}{3/2}\right] = 1\cdot\frac{2}{1}\cdot\frac{2}{3}[/tex]

[tex]R_4 = \frac{W_4}{H_4} = \frac{W_3 + \frac{1}{H_3}}{H_3} = \frac{2 + 2/3}{3/2} = \left[\frac{8/3}{3/2}\right] = 1\cdot\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}[/tex]

[tex]R_5 = \frac{W_5}{H_5} = \frac{W_4}{H_4 + \frac{1}{W_4}} = \frac{8/3}{3/2+3/8}= \left[\frac{8/3}{15/8}\right] = 1\cdot\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}[/tex]

[tex]R_6 = \frac{W_6}{H_6} = \frac{W_5 + \frac{1}{H_5}}{H_5} = \frac{8/3+8/15}{15/8} = \left[\frac{16/5}{15/8}\right] = 1\cdot\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}[/tex]

[tex]R_7 = \frac{W_7}{H_7} = \frac{W_6}{H_6 + \frac{1}{W_6}}= \frac{16/5}{15/8+5/16} = \left[\frac{16/5}{35/16}\right] = 1\cdot\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}[/tex]

By keeping the fractional factors, you see the progression as the Wallis Product, so if you believe the given expansion, you have shown the ratio of the rectangle sides has the [itex]\pi/2[/itex] limit.
 
  • #3
That's definitely it! Thank you very much.
 

Related to Solving the Wallis Product: Constructing Rectangles for $\pi/2$

What is the Wallis Product?

The Wallis Product is a mathematical formula used to approximate the value of pi, denoted as π. It is named after the English mathematician John Wallis who first discovered it in the 17th century.

How does the Wallis Product work?

The Wallis Product involves constructing a sequence of rectangles with specific dimensions and calculating their total area. By taking the limit of this sequence as the number of rectangles approaches infinity, the value of pi can be approximated.

Why is solving the Wallis Product important?

Solving the Wallis Product is important because it provides a simple and efficient way to approximate the value of pi. It is also a fundamental concept in mathematics and can be applied in various fields such as engineering, physics, and computer science.

What are the challenges in solving the Wallis Product?

One of the main challenges in solving the Wallis Product is finding an accurate approximation of pi. This requires a large number of rectangles to be constructed and calculated, which can be time-consuming and computationally intensive.

Are there any real-life applications of the Wallis Product?

Yes, the Wallis Product has several real-life applications. It is used in fields such as architecture, where precise calculations of circular and spherical shapes are needed. It is also used in numerical analysis and in developing algorithms for calculating pi to a desired degree of accuracy.

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