Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 8, Problem 12P

(a)

To determine

To show: That $12xn+xn+1+xn+2+xn+3=2$ , and find a similar equation for the $y$ -coordinates.

(a)

Expert Solution

Answer to Problem 12P

It is proved that $12xn+xn+1+xn+2+xn+3=2$ , and similar equation in terms of $y$ is $12yn+yn+1+yn+2+yn+3=32$ .

Explanation of Solution

Given:

The vertices of the square are $P1(0,1)$ , $P2(1,1)$ , $P3(1,0)$ , and $P4(0,0)$ .

Calculation:

Draw the diagram as follows.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 8, Problem 12P

$Pn(xn,yn)$ is the midpoint of $Pn−4$ and $Pn−3$ . So, $xn=12(xn−4+xn−3)$ for $n≥5$ .

Use the induction on $n$ , to prove the result.

Let $n=1$ ,

Substitute $n=1$ in the equation $12xn+xn+1+xn+2+xn+3=2$ .

$12x1+x1+1+x1+2+x1+3=212x1+x2+x3+x4=212(0)+(1)+(1)+(0)=22=2 (True)$

Assume the result is true for $n=k−1$ .

$12xk−1+xk−1+1+xk−1+2+xk−1+3=212xk−1+xk+xk+1+xk+2=2$

Assume the result is true for $n=k$ .

$12xk+xk+1+xk+2+xk+3=2$

Since $Pk+3(xk+3,yk+3)$ is the midpoint of $Pk+3−4$ , and $Pk+3−3$ .

$12xk+xk+1+xk+2+12(xk+3−4+xk+3−3)=212xk+xk+1+xk+2+12(xk−1+xk)=212xk−1+xk+xk+1+xk+2=2$

Since the result is true for $n=k−1$ .

$12xk−1+xk+xk+1+xk+2=22=2$

Therefore, by the induction, the result $12xn+xn+1+xn+2+xn+3=2$ is true for all $n$ .

$Pn(xn,yn)$ is the midpoint of $Pn−4$ and $Pn−3$ . So, $yn=12(yn−4+yn−3)$ for $n≥5$ .

Consider $12yn+yn+1+yn+2+yn+3$ .

Let $n=1$ ,

Substitute $n=1$ in the equation $12yn+yn+1+yn+2+yn+3$ .

$12yn+yn+1+yn+2+yn+3=12y1+y1+1+y1+2+y1+3=12y1+y2+y3+y4=12(1)+(1)+(0)+(0)=32$

Assume the result is true for $n=k−1$ .

Substitute $n=k−1$ in the equation $12yn+yn+1+yn+2+yn+3$ .

$12yk−1+yk+yk+1+yk+2=32$

Assume the result is true for $n=k$ .

Substitute $n=k$ in the equation $12yn+yn+1+yn+2+yn+3$ .

$12yk+yk+1+yk+2+yk+3=32$

Since $Pk+3(xk+3,yk+3)$ is the midpoint of $Pk+3−4$ , and $Pk+3−3$ .

$12yk+yk+1+yk+2+12(yk+3−4+yk+3−3)=3212yk+yk+1+yk+2+12(yk−1+yk)=3212yk−1+yk+yk+1+yk+2=32$

Since the result is true for $n=k−1$ .

$12yk−1+yk+yk+1+yk+2=3232=32$

Therefore, by the induction, the result $12yn+yn+1+yn+2+yn+3=32$ is true for all $n$ .

Therefore, it is proved that $12xn+xn+1+xn+2+xn+3=2$ , and similar equation in terms of $y$ is $12yn+yn+1+yn+2+yn+3=32$ .

(b)

To determine

To find: The coordinate of $P$ .

(b)

Expert Solution

Answer to Problem 12P

The point $P$ is $(47, 37)$ .

Explanation of Solution

Given:

The vertices of the square are $P1(0,1)$ , $P2(1,1)$ , $P3(1,0)$ , and $P4(0,0)$ .

Calculation:

To find the $x$ coordinate for $P$ , take the limit tends to infinity for the relation $12xn+xn+1+xn+2+xn+3=2$ .

$limn→∞(12xn+xn+1+xn+2+xn+3)=limn→∞(2)limn→∞(12xn)+limn→∞(xn+1)+limn→∞(xn+2)+limn→∞(xn+3)=limn→∞(2)12limn→∞(xn)+limn→∞(xn+1)+limn→∞(xn+2)+limn→∞(xn+3)=2$

Since the values of the above all limits are equal.

$12limn→∞(xn)+limn→∞(xn)+limn→∞(xn)+limn→∞(xn)=2limn→∞(xn)(12+1+1+1)=272limn→∞(xn)=2limn→∞(xn)=47$

To find the $y$ coordinate for $P$ , take the limit tends to infinity for the relation $12yn+yn+1+yn+2+yn+3=32$ .

$limn→∞(12yn+yn+1+yn+2+yn+3)=limn→∞(32)limn→∞(12yn)+limn→∞(yn+1)+limn→∞(yn+2)+limn→∞(yn+3)=limn→∞(32)12limn→∞(yn)+limn→∞(yn+1)+limn→∞(yn+2)+limn→∞(yn+3)=32$

Since the values of the above all limits are equal.

$12limn→∞(yn)+limn→∞(yn)+limn→∞(yn)+limn→∞(yn)=32limn→∞(yn)(12+1+1+1)=3272limn→∞(xn)=32limn→∞(yn)=37$

Therefore, the point $P$ is $(47, 37)$ .

Chapter 8, Problem 11P

Chapter 8, Problem 14P

Chapter 8 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

Ch. 8.1 - Prob. 11E Ch. 8.1 - Prob. 12E Ch. 8.1 - Prob. 13E Ch. 8.1 - Prob. 14E Ch. 8.1 - Prob. 15E Ch. 8.1 - Prob. 16E Ch. 8.1 - Prob. 17E Ch. 8.1 - Prob. 18E Ch. 8.1 - Prob. 19E Ch. 8.1 - Prob. 20E Ch. 8.1 - Prob. 21E Ch. 8.1 - Prob. 22E Ch. 8.1 - Prob. 23E Ch. 8.1 - Prob. 24E Ch. 8.1 - Prob. 25E Ch. 8.1 - Prob. 26E Ch. 8.1 - Prob. 27E Ch. 8.1 - Prob. 28E Ch. 8.1 - Prob. 29E Ch. 8.1 - Prob. 30E Ch. 8.1 - Prob. 31E Ch. 8.1 - Prob. 32E Ch. 8.1 - Prob. 33E Ch. 8.1 - Prob. 34E Ch. 8.1 - Prob. 35E Ch. 8.1 - Prob. 36E Ch. 8.1 - Prob. 37E Ch. 8.1 - Prob. 38E Ch. 8.1 - Prob. 39E Ch. 8.1 - Prob. 40E Ch. 8.1 - Prob. 41E Ch. 8.1 - Prob. 42E Ch. 8.1 - Prob. 43E Ch. 8.1 - Prob. 44E Ch. 8.1 - Prob. 45E Ch. 8.1 - Prob. 46E Ch. 8.1 - Prob. 47E Ch. 8.1 - Prob. 48E Ch. 8.1 - Prob. 49E Ch. 8.1 - Prob. 50E Ch. 8.1 - Prob. 51E Ch. 8.1 - Prob. 52E Ch. 8.1 - Prob. 53E Ch. 8.1 - Prob. 54E Ch. 8.1 - Prob. 55E Ch. 8.1 - Prob. 56E Ch. 8.1 - Prob. 57E Ch. 8.1 - Prob. 58E Ch. 8.1 - Prob. 59E Ch. 8.1 - Prob. 60E Ch. 8.2 - 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That 1 2 x n + x n + 1 + x n + 2 + x n + 3 = 2 , and find a similar equation for the y -coordinates.

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Answer to Problem 12P

Explanation of Solution

To find: The coordinate of P<m:math><m:mi>P</m:mi></m:math> .

Answer to Problem 12P

Explanation of Solution

Chapter 8 Solutions

To find: The coordinate of $P$ .