BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4.7, Problem 26E

(a)

To determine

To find: The root of the equation x3x=1 correct to six decimal places.

Expert Solution

Answer to Problem 26E

The root is 1.324718.

Explanation of Solution

Formula used:

Newton’s method

xn+1=xnf(xn)f(xn)

Given:

f(x)=x3x1 with x1=1

Calculation:

f(x)=x3x1

Differentiate with respect to x ,

f(x)=3x21

xn+1=xnf(xn)f(xn)=xnxn3xn13xn21

x1=1

Substitute n=1 in xn+1=xnxn3xn13xn21

x2=x1x13x113x121

Substitute x1=1 in x2=x1x13x113x121

x2=113113×121=112=1+12=1.5

Substitute n=2 in xn+1=xnxn3xn13xn21

x3=x2x23x213x221

Substitute x2=1.5 in x2=x1x13x113x121

x3=1.51.531.513×1.521=1.347826

Substitute n=3 in xn+1=xnxn3xn13xn21

x4=x3x33x313x321

Substitute x3=1.347826 in x4=x3x33x313x321

x4=1.3478261.34782631.34782613×1.34782621=1.325200

Substitute n=4 in xn+1=xnxn3xn13xn21

x5=x4x43x413x421

Substitute x4=1.325200 in x5=x4x43x413x421

x5=1.3252001.32520031.32520013×1.32520021=1.324718

Substitute n=5 in xn+1=xnxn3xn13xn21

x6=x5x53x513x521

Substitute x5=1.324718 in x6=x5x53x513x521

x6=1.3247181.32471831.32471813×1.32471821=1.324718

x5=x6

Hence the root is 1.324718

(b)

To determine

To find: The root of the equation x3x=1 correct to six decimal places.

Expert Solution

Answer to Problem 26E

The root is 1.324718.

Explanation of Solution

Given:

f(x)=x3x1 with x1=1

Formula used:

Newton’s method

xn+1=xnf(xn)f(xn)

Calculation:

f(x)=x3x1

Differentiate with respect to x ,

f(x)=3x21

xn+1=xnf(xn)f(xn)=xnxn3xn13xn21

x1=0.6

Substitute n=1 in xn+1=xnxn3xn13xn21

x2=x1x13x113x121

Substitute x1=1 in x2=x1x13x113x121

x2=0.60.630.613×0.621=17.9

Substitute n=2 in xn+1=xnxn3xn13xn21

x3=x2x23x213x221

Substitute x2=17.9 in x2=x1x13x113x121

x3=17.917.9317.913×17.921=11.946802

Substitute n=3 in xn+1=xnxn3xn13xn21

x4=x3x33x313x321

Substitute x3=11.946802 in x4=x3x33x313x321

x4=11.94680211.946802311.94680213×11.94680221=7.985520

Substitute n=4 in xn+1=xnxn3xn13xn21

x5=x4x43x413x421

Substitute x4=7.985520 in x5=x4x43x413x421

x5=7.9855207.98552037.98552013×7.98552021=5.356909

Substitute n=5 in xn+1=xnxn3xn13xn21

x6=x5x53x513x521

Substitute x5=5.356909 in x6=x5x53x513x521

x6=5.3569095.35690935.35690913×5.35690921=3.624996

Substitute n=6 in xn+1=xnxn3xn13xn21

x7=x6x63x613x621

Substitute x6=3.624996 in x7=x6x63x613x621

x7=3.6249963.62499633.62499613×3.62499621=2.505589

Substitute n=7 in xn+1=xnxn3xn13xn21

x8=x7x73x713x721

Substitute x7=2.505589 in x8=x7x73x713x721

x8=2.5055892.50558932.50558913×2.50558921=1.820129

Substitute n=8 in xn+1=xnxn3xn13xn21

x9=x8x83x813x821

Substitute x8=1.820129 in x9=x8x83x813x821

x9=1.8201291.82012931.82012913×1.82012921=1.461044

Substitute n=9 in xn+1=xnxn3xn13xn21

x10=x9x93x913x921

Substitute x9=1.461044 in x10=x9x93x913x921

x10=1.4610441.46104431.46104413×1.46104421=1.339323

Substitute n=10 in xn+1=xnxn3xn13xn21

x11=x10x103x1013x1021

Substitute x10=1.339323 in x11=x10x103x1013x1021

x11=1.3393231.33932331.33932313×1.33932321=1.324913

Substitute n=11 in xn+1=xnxn3xn13xn21

x12=x11x113x1113x1121

Substitute x11=1.324913 in x12=x11x113x1113x1121

x12=1.3249131.32491331.32491313×1.32491321=1.324718

Substitute n=12 in xn+1=xnxn3xn13xn21

x13=x12x123x1213x1221

Substitute x12=1.324718 in x13=x12x123x1213x1221

x13=1.3247181.32471831.32471813×1.32471821=1.324718

x12=x13

Hence the root is 1.324718

(c)

To determine

To find: The root of the equation x3x=1 correct to six decimal places.

Expert Solution

Answer to Problem 26E

The root is 1.324718

Explanation of Solution

Given:

f(x)=x3x1 with x1=1

Formula used:

Newton’s method

xn+1=xnf(xn)f(xn)

Calculation:

f(x)=x3x1

Differentiate with respect to x ,

f(x)=3x21

xn+1=xnf(xn)f(xn)=xnxn3xn13xn21

x1=0.57

Substitute n=1 in xn+1=xnxn3xn13xn21

x2=x1x13x113x121

Substitute x1=0.57 in x2=x1x13x113x121

x2=0.570.5730.5713×0.5721=54.165455

Substitute n=2 in xn+1=xnxn3xn13xn21

x3=x2x23x213x221

Substitute x2=54.165455 in x2=x1x13x113x121

x3=54.16545554.1654553+54.16545513×54.16545521=36.114293

Substitute n=3 in xn+1=xnxn3xn13xn21

x4=x3x33x313x321

Substitute x3=36.114293 in x4=x3x33x313x321

x4=36.11429336.1142933+36.11429313×36.11429321=24.082094

Substitute n=4 in xn+1=xnxn3xn13xn21

x5=x4x43x413x421

Substitute x4=24.082094 in x5=x4x43x413x421

x5=24.08209424.0820943+24.08209413×24.08209421=16.063387

Substitute n=5 in xn+1=xnxn3xn13xn21

x6=x5x53x513x521

Substitute x5=16.063387 in x6=x5x53x513x521

x6=16.06338716.0633873+16.06338713×16.06338721=10.721483

Substitute n=6 in xn+1=xnxn3xn13xn21

x7=x6x63x613x621

Substitute x6=10.721483 in x7=x6x63x613x621

x7=10.72148310.7214833+10.72148313×10.72148321=7.165534

Substitute n=7 in xn+1=xnxn3xn13xn21

x8=x7x73x713x721

Substitute x7=7.165534 in x8=x7x73x713x721

x8=7.1655347.1655343+7.165534913×7.16553421=4.801704

Substitute n=8 in xn+1=xnxn3xn13xn21

x9=x8x83x813x821

Substitute x8=4.801704 in x9=x8x83x813x821

x9=4.8017044.8017043+4.80170413×4.80170421=3.233425

Substitute n=9 in xn+1=xnxn3xn13xn21

x10=x9x93x913x921

Substitute x9=3.233425 in x10=x9x93x913x921

x10=3.2334253.2334253+3.23342513×3.23342521=2.193674

Substitute n=10 in xn+1=xnxn3xn13xn21

x11=x10x103x1013x1021

Substitute x10=2.193674 in x11=x10x103x1013x1021

x11=2.1936742.1936743+2.19367413×2.19367421=1.496867

Substitute n=11 in xn+1=xnxn3xn13xn21

x12=x11x113x1113x1121

Substitute x11=1.496867 in x12=x11x113x1113x1121

x12=1.4968671.4968673+1.49686713×1.49686721=0.997546

Substitute n=12 in xn+1=xnxn3xn13xn21

x13=x12x123x1213x1221

Substitute x12=0.997546 in x13=x12x123x1213x1221

x13=0.9975460.9975463+0.99754613×0.99754621=0.496305

Substitute n=13 in xn+1=xnxn3xn13xn21

x14=x13x133x1313x1321

Substitute x13=0.496305 in x14=x13x133x1313x1321

x14=0.4963050.4963053+0.49630513×0.49630521=2.894162

Substitute n=14 in xn+1=xnxn3xn13xn21

x15=x14x143x1413x1421

Substitute x14=2.894162 in x15=x14x143x1413x1421

x15=2.8941622.8941623+2.89416213×2.89416221=1.967962

Substitute n=15 in xn+1=xnxn3xn13xn21

x16=x15x153x1513x1521

Substitute x15=1.967962 in x16=x15x153x1513x1521

x16=1.9679621.9679623+1.96796213×1.96796221=1.341355

Substitute n=16 in xn+1=xnxn3xn13xn21

x17=x16x163x1613x1621

Substitute x16=1.341355 in x17=x16x163x1613x1621

x17=1.3413551.3413553+1.34135513×1.34135521=0.870187

Substitute n=17 in xn+1=xnxn3xn13xn21

x18=x17x173x1713x1721

Substitute x17=0.870187 in x18=x17x173x1713x1721

x18=0.8701870.8701873+0.87018713×0.87018721=0.249949

Substitute n=18 in xn+1=xnxn3xn13xn21

x19=x18x183x1813x1821

Substitute x18=0.249949 in x19=x18x183x1813x1821

x19=0.2499490.2499493+0.24994913×0.24994921=1.192219

Substitute n=19 in xn+1=xnxn3xn13xn21

x20=x19x193x1913x1921

Substitute x19=1.192219 in x20=x19x193x1913x1921

x20=1.1922191.1922193+1.19221913×1.19221921=0.731952

Substitute n=20 in xn+1=xnxn3xn13xn21

x21=x20x203x2013x2021

Substitute x20=0.731952 in x21=x20x203x2013x2021

x21=0.7319520.7319523+0.73195213×0.73195221=0.355213

Substitute n=21 in xn+1=xnxn3xn13xn21

x22=x21x213x2113x2121

Substitute x21=0.355213 in x22=x21x213x2113x2121

x22=0.3552130.35521330.35521313×0.35521321=1.753322

Substitute n=22 in xn+1=xnxn3xn13xn21

x23=x22x223x2213x2221

Substitute x22=1.753322 in x23=x22x223x2213x2221

x23=(1.753322)(1.753322)3(1.753322)13(1.753322)21=1.189420

Substitute n=23 in xn+1=xnxn3xn13xn21

x24=x23x233x2313x2321

Substitute x23=1.189420 in x24=x23x233x2313x2321

x24=(1.189420)(1.189420)3(1.189420)13(1.189420)21=0.729123

Substitute n=24 in xn+1=xnxn3xn13xn21

x25=x24x243x2413x2421

Substitute x24=0.729123 in x25=x24x243x2413x2421,

x25=(0.729123)(0.729123)3(0.729123)13(0.729123)21=0.377844

Substitute n=25 in xn+1=xnxn3xn13xn21

x26=x25x253x2513x2521

Substitute x25=0.377844 in x26=x25x253x2513x2521,

x26=(0.377844)(0.377844)3(0.377844)13(0.377844)21=1.937872

Substitute n=26 in xn+1=xnxn3xn13xn21

x27=x26x263x2613x2621

Substitute x26=1.937872 in x27=x26x263x2613x2621,

x27=(1.937872)(1.937872)3(1.937872)13(1.937872)21=1.320350

Substitute n=27 in xn+1=xnxn3xn13xn21

x28=x27x273x2713x2721

Substitute x27=1.320350 in x28=x27x273x2713x2721,

x28=(1.320350)(1.320350)3(1.320350)13(1.320350)21=0.851919

Substitute n=28 in xn+1=xnxn3xn13xn21

x29=x28x283x2813x2821

Substitute x28=0.851919 in x29=x28x283x2813x2821,

x29=(0.851919)(0.851919)3(0.851919)13(0.851919)21=0.200959

Substitute n=29 in xn+1=xnxn3xn13xn21

x30=x29x293x2913x2921

Substitute x29=0.200959 in x30=x29x293x2913x2921,

Substitute n=30 in xn+1=xnxn3xn13xn21

x31=x30x303x3013x3021

Substitute x30=1.119386 in x31=x30x303x3013x3021

x31=1.1193861.1193863+1.11938613×1.11938621=0.654291

Substitute n=31 in xn+1=xnxn3xn13xn21

x32=x31x313x3113x3121

Substitute x31=0.654291 in x32=x31x313x3113x3121

x32=0.6542910.6542913+0.65429113×0.65429121=1.547009

Substitute n=32 in xn+1=xnxn3xn13xn21

x33=x32x323x3213x3221

Substitute x32=1.547009 in x33=x32x323x3213x3221

Substitute n=33 in xn+1=xnxn3xn13xn21

x34=x33x333x3313x3321

Substitute x33=1.360050 in x34=x33x333x3313x3321

x34=1.3600501.36005031.36005013×1.36005021=1.325828

Substitute n=34 in xn+1=xnxn3xn13xn21

x35=x34x343x3413x3421

Substitute x34=1.325828 in x35=x34x343x3413x3421

x35=1.3258281.32582831.32582813×1.32582821=1.324719

Substitute n=35 in xn+1=xnxn3xn13xn21

x36=x35x353x3513x3521

Substitute x35=1.324719 in x36=x35x353x3513x3521

Substitute n=36 in xn+1=xnxn3xn13xn21

x37=x36x363x3613x3621

Substitute x36=1.324718 in x37=x36x363x3613x3621

x37=1.3247181.32471831.32471813×1.32471821=1.324718

Here, x36=x37

Thus, the root of x is 1.324718.

(d)

To determine

To sketch: the graph f(x)=x3x1 and its tangent lines at 1, 0.6, 0.57 to explain the dependency of initial approximation in Newton’s method.

Expert Solution

Explanation of Solution

Given:

f(x)=x3x1

Calculation:

f(x)=x3x1

The graph is,

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 26E

In Figure 1,

The tangent line corresponding to x1=1 results in a sequence of approximations that converges quite quickly.

The tangent line corresponding to x1=0.6 is close to being horizontal, so x2 is quite far from the root. But the sequence converges slowly.

The tangent line corresponding to x1=0.57 is nearly horizontal, x2 is farther away from the root, and the sequence takes more iterations to converge.

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