Numerical Analysis
Numerical Analysis
3rd Edition
ISBN: 9780134696454
Author: Sauer, Tim
Publisher: Pearson,
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Chapter 1.5, Problem 1E

a.

To determine

To calculate:To apply two steps of the secant method to the equation x3=2x+2 with initial guesses x0=1andx1=2 .

a.

Expert Solution
Check Mark

Answer to Problem 1E

Thus, x2=85 and x3=1.742268041 .

Given information:Initial guesses x0=1andx1=2 and equation is (a)x3=2x+2 .

Calculation:

With the initial guesses,  x0=1andx1=2 .

Now, using the formula, we will get the following.

   xi+1=xif1(xi)(xixi1)f1(xi)f1(xi1)=xi(xi32xi2)(xixi1)(xi32xi2)(xi132xi12)=xi(xi32xi2)(xixi1)xi32xixi13+2xi1

Let us find the value of  x2 .

   x2=x1(x132x12)(x1x0)x132x1x03+2x0=2(232(2)2)(21)232(2)13+2(1)

   =225=85

Let us find the value of  x3 .

   x3=x2(x232x22)(x2x1)x232x2x13+2x1=85((85)32(85)2)(852)(85)32(85)23+2(2)=1.742268041

Explanation of Solution

Given information:Initial guesses x0=1andx1=2 and equation is (a)x3=2x+2 .

Calculation:

With the initial guesses,  x0=1andx1=2 .

Now, using the formula, we will get the following.

   xi+1=xif1(xi)(xixi1)f1(xi)f1(xi1)=xi(xi32xi2)(xixi1)(xi32xi2)(xi132xi12)=xi(xi32xi2)(xixi1)xi32xixi13+2xi1

Let us find the value of  x2 .

   x2=x1(x132x12)(x1x0)x132x1x03+2x0=2(232(2)2)(21)232(2)13+2(1)

   =225=85

Let us find the value of  x3 .

   x3=x2(x232x22)(x2x1)x232x2x13+2x1=85((85)32(85)2)(852)(85)32(85)23+2(2)=1.742268041

b.

To determine

To calculate:To apply two steps of the secant method to the equation ex+x=7 with initial guesses x0=1andx1=2 .

b.

Expert Solution
Check Mark

Answer to Problem 1E

Thus, x2=1.578707248 and x3=1.660160100

Given information:Initial guesses x0=1andx1=2 and equation is (b)ex+x=7 .

Calculation:

Let us consider the following equation.

   ex+x=7ex+x7=0

Or consider the following function

   f2(x)=ex+x7

With the initial guesses,  x0=1  and x1=2 .

Now, using the formula, we will get the following.

   xi+1=xif2(xi)(xixi1)f2(xi)f2(xi1)=xi(exi+xi7)(xixi1)(exi+xi7)(ex4+xi17)=xi(exi+xi7)(xixi1)exi+xiexh1

Let us find the value of  x2 .

   x2=x1(ex1+x17)(x1x0)ex1+x1ex0x0=2(e2+27)(21)e2+2e11=2e25e2e1+1=1.578707248

Let us find the value of  x3 .

   x3=x2(ex2+x27)(x2x1)ex1+x1ex0x0=1.578707248(e1.580n248+1.5787072487)(1.5787072482)e1.5800n48+1.578707248e22=1.660160100

c.

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1andx1=2 .

Thus, x2=1.092906580 and x3=1.119356686

Given information:Initial guesses x0=1andx1=2 and equation is (c)ex+sinx=4

Calculation:

Let us consider the following equation.

   ex+sin(x)=4ex+sin(x)4=0

Or consider the following function.

   f3(x)=ex+sin(x)4

With the initial guesses,  x0=1  and x1=2 .

Now, using the formula, we will get the following.

   xi+1=xif3(xi)(xixi1)f3(xi)f3(xi1)=xi(ex1+sin(xi)4)(xixi1)(ex1+sin(xi)4)(ex1+1+sin(xi1)4)=xi(exi+sin(xi)4)(xixi1)exi+sin(xi)exhsin(xi1)

Let us find the value of  x2 .

   x2=x1(ex1+sin(x1)4)(x1x0)ex1+sin(x1)ex0sin(x0)=2(e2+sin(2)4)(21)e2+sin(2)e1sin(1)=2e2+sin(2)4e2e1+sin(2)sin(1)=1.092906580

Let us find the value of  x3 .

   x3=x2(ex2+sin(x2)4)(x2x1)ex2+sin(x2)ex1sin(x1)=1.092906580(e1.09906550+sin(1.092906580)4)(1.0929065802)e1.092906580+sin(1.092906580)e2sin(2)=2e2+sin(2)4e2e1+sin(2)sin(1)=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1andx1=2 and equation is (b)ex+x=7 .

Calculation:

Let us consider the following equation.

   ex+x=7ex+x7=0

Or consider the following function

   f2(x)=ex+x7

With the initial guesses,  x0=1  and x1=2 .

Now, using the formula, we will get the following.

   xi+1=xif2(xi)(xixi1)f2(xi)f2(xi1)=xi(exi+xi7)(xixi1)(exi+xi7)(ex4+xi17)=xi(exi+xi7)(xixi1)exi+xiexh1

Let us find the value of  x2 .

   x2=x1(ex1+x17)(x1x0)ex1+x1ex0x0=2(e2+27)(21)e2+2e11=2e25e2e1+1=1.578707248

Let us find the value of  x3 .

   x3=x2(ex2+x27)(x2x1)ex1+x1ex0x0=1.578707248(e1.580n248+1.5787072487)(1.5787072482)e1.5800n48+1.578707248e22=1.660160100

c.

To determine

To calculate:To apply two steps of the secant method to the equation ex+sinx=4 with initial guesses x0=1andx1=2 .

c.

Expert Solution
Check Mark

Answer to Problem 1E

Thus, x2=1.092906580 and x3=1.119356686

Explanation of Solution

Given information:Initial guesses x0=1andx1=2 and equation is (c)ex+sinx=4

Calculation:

Let us consider the following equation.

   ex+sin(x)=4ex+sin(x)4=0

Or consider the following function.

   f3(x)=ex+sin(x)4

With the initial guesses,  x0=1  and x1=2 .

Now, using the formula, we will get the following.

   xi+1=xif3(xi)(xixi1)f3(xi)f3(xi1)=xi(ex1+sin(xi)4)(xixi1)(ex1+sin(xi)4)(ex1+1+sin(xi1)4)=xi(exi+sin(xi)4)(xixi1)exi+sin(xi)exhsin(xi1)

Let us find the value of  x2 .

   x2=x1(ex1+sin(x1)4)(x1x0)ex1+sin(x1)ex0sin(x0)=2(e2+sin(2)4)(21)e2+sin(2)e1sin(1)=2e2+sin(2)4e2e1+sin(2)sin(1)=1.092906580

Let us find the value of  x3 .

   x3=x2(ex2+sin(x2)4)(x2x1)ex2+sin(x2)ex1sin(x1)=1.092906580(e1.09906550+sin(1.092906580)4)(1.0929065802)e1.092906580+sin(1.092906580)e2sin(2)=2e2+sin(2)4e2e1+sin(2)sin(1)=1.119356686

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Chapter 1 Solutions

Numerical Analysis

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