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...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Videos

Question

Chapter 11.2, Problem 70E

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 69E

Chapter 11.2, Problem 71E

Students have asked these similar questions

3.1 Limits 1. If lim f(x)=-6 and lim f(x)=5, then lim f(x). Explain your choice. x+3° x+3* x+3 (a) Is 5 (c) Does not exist (b) is 6 (d) is infinite

1 pts Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is Question 1 -0.246 0.072 -0.934 0.478 -0.914 -0.855 0.710 0.262 .

2. Answer the following questions. (A) [50%] Given the vector field F(x, y, z) = (x²y, e", yz²), verify the differential identity Vx (VF) V(V •F) - V²F (B) [50%] Remark. You are confined to use the differential identities. Let u and v be scalar fields, and F be a vector field given by F = (Vu) x (Vv) (i) Show that F is solenoidal (or incompressible). (ii) Show that G = (uvv – vVu) is a vector potential for F.

Chapter 11 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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