Question

Let q>0 be a real constant. Instead of using square boxes as in Question 3, imagine boxes in the shape of rectangles, where box n has base width 1/nᵠ and height 1/n. (Treat Question 3 as the special case q=1.) (a) Imagine packing these new rectangular boxes according to the process detailed in Question 3. The picture from 3(a) won't change much, so do nothing here. (b) Upgrade your formula from 3(b) to find the interval [a ₙ,bₙ ] supporting box number 2 ⁿ . The result should apply for any n≥0 and any q>0. (c) Adapt your work in 3(c) to apply with a general q>0. In particular, show that the height-limit of 1 remains correct. (d) Express as a series the total width of shelf required to hold all the boxes. Then determine which values of q>0 correspond to a finite total width. (e) The box-packing discussion above proves convergence for a certain family of series. Identify this family and make intelligent remarks that relate your findings here with important results discussed in class.

Previous question

Answer 1

To find the interval [aₙ, bₙ] supporting box number 2ⁿ, substitute n=2ⁿ into the formulas for base width and height. To adapt the work for a general q>0, replace the exponent of n in the base width formula with q. Express the total width of the shelf as a series by summing the base widths of all the boxes. Convergence of the series depends on the value of q.

Step-by-step explanation:

To find the interval [aₙ,bₙ] supporting box number 2ⁿ, we need to determine the width and height of this box. Given that box n has base width 1/nᵠ and height 1/n, we can substitute n=2ⁿ into these formulas. The base width of box 2ⁿ will be 1/(2ⁿ)ᵠ = 1/(2ⁿᵠ), and the height will be 1/(2ⁿ). Therefore, the interval [aₙ,bₙ] supporting box number 2ⁿ is [1/(2ⁿᵠ), 1/(2ⁿ)].

To adapt the work in 3(c) to apply with a general q>0, we can use the same formulas as before, but replace the exponent of n in the base width with q. The base width of box n with exponent q will be 1/nᵠ and the height will be 1/n. The height limit of 1 remains correct because it is not affected by the value of q.

To express the total width of the shelf required to hold all the boxes as a series, we need to sum the base widths of all the boxes. Using the formula for the base width of box n with exponent q as 1/nᵠ, the series for the total width is ∑(1/nᵠ) from n=1 to infinity. The series will have a finite total width when it converges. Convergence depends on the value of q, and values of q>0 that result in the series converging correspond to a finite total width.

Answer 2

In part (b), the interval [aₙ,bₙ] supporting box number 2ⁿ is [1/(2ⁿᵠ+1), 1/(2ⁿ+1)]. In part (c), the height-limit of 1 remains correct for a general q>0. In part (d), the total width of the shelf required to hold all the boxes can be expressed as a series ∑(1/(nᵠ+1)), which is finite when q>1.

Step-by-step explanation:

Let's address each part of the question:

(b) To find the interval [aₙ,bₙ] supporting box number 2ⁿ, we need to update the formula from Question 3(b). The new formula for the base width is 1/(nᵠ+1) and the new formula for the height is 1/(n+1). So, the interval would be [1/(2ⁿᵠ+1), 1/(2ⁿ+1)].

(c) To adapt the work in Question 3(c) for a general q>0, we need to find the height-limit of 1. Since the height is 1/(n+1), as n approaches infinity, the height approaches 0. So, the height-limit of 1 remains correct.

(d) To express the total width of shelf required to hold all the boxes as a series, we can use the formula for the base width and sum it over all values of n. The series would be ∑(1/(nᵠ+1)). The total width is finite if the series converges, which happens when q>1.

(e) The box-packing discussion above proves convergence for a family of series called p-series, which are of the form ∑(1/n^p). This is related to the p-series test, which states that if p>1, then the series converges.