Explanation Given: Length of the table is L Calculation: (a) After L 2 stage the total amount of table removed is L 4 + 2 L 4 2 + 4 L 4 3 + 8 L 4 4 + ... + 2 N − 1 L 4 N = L ( 1 4 + 2 4 2 + 4 4 3 + ... + 2 N − 1 4 N ) = L ( 1 4 + 1 8 + 1 16 + ... + 1 2 N + 1 ) At 1 stage there are two remaining sections each of length L − L 4 2 = 3 L 8 < L 2 Suppose at K t h each of the 2 K remaining section has length less than L 2 K The ( K + 1 ) stage is obtained by removing the section of length L 4 K + 1 centered at the mid point of each segment in the K t h stage. Let a K and a K + 1 represent the length of each segment in K t h and ( K + 1 ) t h stage respectively...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

See similar textbooks

Videos

Question

Chapter 11.2, Problem 64E

To determine

(a) We need to show that at $N^{t h}$ stage, each remaining section has length less than $\frac{L}{2^{N}}$ and that the total amount of table removed is $L (\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ... + \frac{1}{2^{N + 1}})$

(b) We need to show that as the limit $N \to \infty$ preciously one half of the table remains.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.2, Problem 63E

Chapter 11.2, Problem 65E

Students have asked these similar questions

For each given function f(x) find f'(x) using the rules learned in section 9.5. 1. f(x)=x32 32x 2. f(x)=7x+13 3. f(x) = x4 4. f(x) = √√x³ 5. f(x) = 3x²+ 3 x2

Find: lim x →-6 f (x) limx-4 f (x) lim x-1 f (x) lim x →4 f (x) (-6,3) • (-1,5) -8 -7 (-6,-2) 4+ (4,5) (4,2) • (-1,1) -6

3 2 Find: ƒ(1) lim f(x) 14-x 2 ƒ(2) lim f(x) x-2- lim f(x) x+2+ lim f(x) x→4 3 y=f(x)

Chapter 11 Solutions

Calculus - Standalone book

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

(a) We need to show that at N t h stage, each remaining section has length less than L 2 N and that the total amount of table removed is L ( 1 4 + 1 8 + 1 16 + ... + 1 2 N + 1 ) (b) We need to show that as the limit N → ∞ preciously one half of the table remains.

Solution Summary: The author explains that at N t h stage, each remaining section has length less than L 2 N and that the total amount of table removed is L.

Videos

Want to see the full answer?

Chapter 11 Solutions

(a) We need to show that at N t h stage, each remaining section has length less than L 2 N and that the total amount of table removed is L ( 1 4 + 1 8 + 1 16 + ... + 1 2 N + 1 ) (b) We need to show that as the limit N → ∞ preciously one half of the table remains.

Solution Summary: The author explains that at N t h stage, each remaining section has length less than L 2 N and that the total amount of table removed is L.

Videos

(a) We need to show that at Nth stage, each remaining section has length less than L2N and that the total amount of table removed is L(14+18+116+...+12N+1) (b) We need to show that as the limit N→∞ preciously one half of the table remains.

Want to see the full answer?

Chapter 11 Solutions