Explanation Given: x 3 + 2 x − 4 = 0 and x 1 = 1 Concept Used: Newton’s method to find ( n + 1 ) th approximate root of an equation f ( x ) = 0 is x n + 1 = x n − f ( x n ) f ′ ( x n ) Where, x n is nth approximate root and f ′ ( x n ) ≠ 0 Calculation: To find third approximate root of equation x 3 + 2 x − 4 = 0 where x 1 = 1 Use Newton’s method, that states ( n + 1 ) th approximate root of an equation f ( x ) = 0 is x n + 1 = x n − f ( x n ) f ′ ( x n ) Here, f ( x ) = x 3 + 2 x − 4 , and x 1 = 1 Now, calculate f ′ ( x ) as f ′ ( x ) = d d x ( x 3 + 2 x − 4 ) = d d x x 3 + 2 d d x x − d d x 4 = 3 x 2 + 2 − 0 = 3 x 2

15th Edition

ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Chapter 3.8, Problem 21E

To determine

To find third approximate root of equation

x3+2x−4=0

Using Newton’s method, where x1=1

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Chapter 3.8, Problem 20E

Chapter 3.8, Problem 22E

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Could you explain this using the formula I attached and polar coorindates

Could you explain this using the formula I attached and polar coordinates

Chapter 3 Solutions

Biocalculus

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To find third approximate root of equation x 3 + 2 x − 4 = 0 Using Newton’s method, where x 1 = 1

Solution Summary: The author explains Newton’s method to find the third approximate root of an equation x3+2x-4=0.

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To find third approximate root of equation

x3+2x−4=0

Using Newton’s method, where x1=1

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Chapter 3 Solutions

To find third approximate root of equation x 3 + 2 x − 4 = 0 Using Newton’s method, where x 1 = 1

Solution Summary: The author explains Newton’s method to find the third approximate root of an equation x3+2x-4=0.

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Videos

To find third approximate root of equation x3+2x−4=0 Using Newton’s method, where x1=1

Want to see the full answer?

Chapter 3 Solutions

To find third approximate root of equation

x3+2x−4=0

Using Newton’s method, where x1=1