# Difference between Bisection Method and Newton Raphson Method

Numerical methods are the set of tasks by applying arithmetic operations to numerical equations. We can formulate mathematical problems to find the approximate result. This formulation is called the numerical implementation of the problem. In this, there is no need for algorithms. Programming logic is then developed for numerical implementation. The programming is usually done with some high-level languages like Fortran, Basic, etc.

### Bisection Method

This method is based on the repeated application of the intermediate value property. Let f(x) is continuous function in the closed interval [x_{1}, x_{2}], if f(x_{1}), f(x_{2}) are of opposite signs, then there is at least one root α in the interval (x_{1}, x_{2}), such that f(α) = 0.

**Formula is: **x_{2 }= (x_{0 }+ x_{1}) / 2

### Newton Raphson Method

The Newton Raphson Method is the process for the determination of a real root of an equation f(x)=0 given just one point close to the desired root.

**Formula is: **x_{1} = x_{0} – f(x_{0})/f'(x_{0})

### Comparison between Bisection Method and Newton Raphson Method

Sr. No. | Bisection Method | Newton Raphson Method |
---|---|---|

1. | In the Bisection Method, the rate of convergence is linear thus it is slow. | In the Newton Raphson method, the rate of convergence is second-order or quadratic. |

2. |
In Bisection Method we used following formula x |
In Newton Raphson method we used following formula x |

3. | In this method, we take two initial approximations of the root in which the root is expected to lie. | In this method, we take one initial approximation of the root. |

4. | The computation of function per iteration is 1. | The computation of function per iteration is 2. |

5. | The initial approximation is less sensitive. | The initial approximation is very sensitive. |

6. | In the Bisection Method, there is no need to find derivatives. | In the Newton Raphson method, there is a need to find derivatives. |

7. | This method is not applicable for finding complex, multiple, and nearly equal two roots. | This method is applicable for finding complex, multiple, and nearly equal two roots. |

**Question 1: Find a root of an equation f(x) = x ^{3 }– x – 1 **

**Solution:**

Given equation f(x) = x

^{3 }– x – 1let x = 0, 1, 2

In 1st iteration:f(1) = -1 < 0 and f(2) = 5 > 0

The root lies between these two points 1 and 2

x

_{0 }= 1 + 2/2 = 1.5f(x

_{0}) = f(1.5) = 0.875 > 0

In 2nd iteration:f(1) = -1 < 0 and f(1.5) = 0.875 > 0

The root lies between these two points 1 and 1.5

x

_{1 }= 1 + 1.5/2 = 1.25f(x

_{1}) = f(1.25) = -0.29688 < 0

In 3rd iteration:f(1.25) = -0.29688 < 0 and f(1.5) = 0.875 > 0

The root lies between these two points 1.25 and 1.5

x

_{2 }= 1.25 + 1.5/2 = 1.375f(x

_{2}) = f(1.375) = 0.22461 > 0

In 4th iteration:f(1.25) = -0.29688 < 0 and f(1.375) = 0.22461 > 0

The root lies between these two points 1.25 and 1.375

x

_{3 }= 1.25 + 1.375/2 = 1.3125f(x

_{3}) = f(1.3125) = -0.05151 < 0

In 5th iteration:f(1.3125) = -0.05151 < 0 and f(1.375) = 0.22461 > 0

The root lies between these two points 1.3125 and 1.375

x

_{4 }= 1.3125 + 1.375/2 = 1.34375f(x

_{4}) = f(1.34375) = 0.08261 > 0

In 6th iteration:f(1.3125) = -0.05151 < 0 and f(1.34375) = 0.08261 > 0

The root lies between these two points 1.3125 and 1.34375

x

_{5 }= 1.3125 + 1.34375/2 = 1.32812f(x

_{5}) = f(1.32812) = 0.01458 > 0

In 7th iteration:f(1.3125) = -0.05151 < 0 and f(1.32812) = 0.01458 > 0

The root lies between these two points 1.3125 and 1.32812

x

_{6 }= 1.3125 + 1.32812/2 = 1.32031f(x

_{6}) = f(1.32031) = -0.01871 < 0

In 8th iteration:f(1.32031) = -0.01871 < 0 and f(1.32812) = 0.01458 > 0

The root lies between these two points 1.32031 and 1.32812

x

_{7 }= 1.32031 + 1.32812/2 = 1.32422f(x

_{7}) = f(1.32422) = -0.00213 < 0

In 9th iteration:f(1.32422) = -0.00213 < 0 and f(1.32812) = 0.01458 > 0

The root lies between these two points 1.32422 and 1.32812

x

_{8 }= 1.32422 + 1.32812/2 = 1.32617f(x

_{8}) = f(1.32617) = 0.00621 > 0

In 10th iteration:f(1.32422) = -0.00213 < 0 and f(1.32617) = 0.00621 > 0

The root lies between these two points 1.32422 and 1.32617

x

_{9 }= 1.32422 + 1.32617/2 = 1.3252f(x

_{9}) = f(1.3252) = 0.00204 > 0

In 11th iteration:f(1.32422) = -0.00213 < 0 and f(1.3252) = 0.00204 > 0

The root lies between these two points 1.32422 and 1.3252

x

_{10 }= 1.32422 + 1.3252/2 = 1.32471f(x

_{10}) = f(1.32471) = -0.00005 < 0

The approximate root of the equation x^{3 }– x – 1 = 0 using the Bisection method is 1.32471

**Question 2: Find a root of an equation f(x) = 2x ^{3 }– 2x – 5**

**Solution:**

Given Equation f(x) = 2x

^{3 }– 2x – 5

In 1st iteration:f(1) = -5 < 0 and f(2) = 7 > 0

The root lies between these two points 1 and 2

x

_{0 }= 1 + 2/2 = 1.5f(x

_{0}) = f(1.5) = 2 × 1.53 – 2 × 1.5 – 5 = -1.25 < 0

In 2nd iteration:f(1.5) = -1.25 < 0 and f(2) = 7 > 0

The root lies between these two points 1.5 and 2

x

_{1 }= 1.5 + 2/2 = 1.75f(x

_{1}) = f(1.75) = 2 × 1.753 – 2 × 1.75 – 5 = 2.21875 > 0

In 3rd iteration:f(1.5) = -1.25 < 0 and f(1.75) = 2.21875 > 0

The root lies between these two points1.5 and 1.75

x

_{2 }= 1.5 + 1.75/2 = 1.625f(x

_{2}) = f(1.625) = 2 × 1.6253 – 2 × 1.625 – 5 = 0.33203 > 0

In 4th iteration:f(1.5) = -1.25 < 0 and f(1.625) = 0.33203 > 0

The root lies between these two points 1.5 and 1.625

x

_{3 }= 1.5 + 1.625/2 = 1.5625f(x

_{3}) = f(1.5625) = 2 × 1.56253 – 2 × 1.5625 – 5 = -0.49561 < 0

In 5th iteration:f(1.5625) = -0.49561 < 0 and f(1.625) = 0.33203 > 0

The root lies between these two points 1.5625 and 1.625

x

_{4 }= 1.5625 + 1.625/2 = 1.59375f(x

_{4}) = f(1.59375) = 2 × 1.593753 – 2 × 1.59375 – 5 = -0.09113 < 0

In 6th iteration:f(1.59375) = -0.09113 < 0 and f(1.625) = 0.33203 > 0

The root lies between these two points 1.59375 and 1.625

x

_{5 }= 1.59375 + 1.625/2 = 1.60938f(x

_{5}) = f(1.60938) = 2 × 1.609383 – 2 × 1.60938 – 5 = 0.1181 > 0

In 7th iteration:f(1.59375) = -0.09113 < 0 and f(1.60938) = 0.1181 > 0

The root lies between these two points 1.59375 and 1.60938

x

_{6 }= 1.59375 + 1.60938/2 = 1.60156f(x

_{6}) = f(1.60156) = 2 × 1.601563 – 2 × 1.60156 – 5 = 0.0129 > 0

In 8th iteration:f(1.59375) = -0.09113 < 0 and f(1.60156) = 0.0129 > 0

The root lies between these two points 1.59375 and 1.60156

x

_{7 }= 1.59375 + 1.60156/2 = 1.59766f(x

_{7}) = f(1.59766) = 2 × 1.597663 – 2 × 1.59766 – 5 = -0.03926 < 0

In 9th iteration:f(1.59766) = -0.03926 < 0 and f(1.60156) = 0.0129 > 0

The root lies between these two points 1.59766 and 1.60156

x

_{8 }= 1.59766 + 1.60156/2 = 1.59961f(x

_{8}) = f(1.59961) = 2 × 1.599613 – 2 × 1.59961 – 5 = -0.01322 < 0

In 10th iteration:Here f(1.59961) = -0.01322 < 0 and f(1.60156) = 0.0129 > 0

The root lies between these two points 1.59961 and 1.60156

x

_{9 }= 1.59961 + 1.60156/2 = 1.60059f(x

_{9}) = f(1.60059) = 2 × 1.600593 – 2 × 1.60059 – 5 = -0.00017 < 0

The Approximate root of the equation 2x^{3 }– 2x – 5 = 0 using Bisection method is 1.60059

**Question 3: Find a root of an equation f(x) = x ^{3 }– x – 1**

**Solution:**

Given equation x

^{3 }– x – 1 = 0Using differentiate method the equation is

∴ f′(x) = 3x

^{2 }– 1Here f(1) = -1 < 0 and f(2) = 5 > 0

∴ Root lies between 1 and 2

x

_{0 }= 1 + 2/ 2 = 1.5

In 1st iteration:f(x

_{0}) = f(1.5) = 0.875f′(x

_{0}) = f′(1.5) = 5.75x

_{1}= x_{0 }– f(x_{0}) / f′(x_{0})x

_{1 }= 1.5 – 0.875/ 5.75x

_{1 }= 1.34783

In 2nd iteration:f(x

_{1}) = f(1.34783) = 0.10068f′(x

_{1}) = f′(1.34783) = 4.44991x

_{2 }= x_{1 }– f(x_{1})/f′(x_{1})x

_{2 }= 1.34783 – 0.10068/4.44991x

_{2 }= 1.3252

In 3rd iteration:f(x

_{2}) = f(1.3252) = 0.00206f′(x

_{2}) = f′(1.3252) = 4.26847x

_{3 }= x_{2 }– f(x_{2})/f′(x_{2})x

_{3 }= 1.3252 – 0.00206/4.26847x

_{3 }= 1.32472

In 4th iteration:f(x

_{3}) = f(1.32472) = 0f′(x

_{3}) = f′(1.32472) = 4.26463x

_{4 }= x_{3 }– f(x_{3})/f′(x_{3})x

_{4 }= 1.32472 – 0/ 4.26463x

_{4 }= 1.32472

The Approximate root of the equation x^{3 }– x – 1 = 0 using the Newton Raphson method is 1.32472

**Question 4: Find a root of an equation f(x) = 2x ^{3 }– 2x – 5**

**Solution:**

Given equation 2x

^{3 }– 2x – 5 = 0Using differentiate method the equation is

∴ f′(x) = 6x

^{2 }– 2Here f(1) = -5 < 0 and f(2) = 7 > 0

∴ Root lies between 1 and 2

x

_{0 }= 1 + 2/ 2 = 1.5

In 1st iteration:f(x

_{0}) = f(1.5) = 2 × 1.53 – 2 × 1.5 – 5 = -1.25f′(x

_{0}) = f′(1.5) = 6 × 1.52 – 2 = 11.5x

_{1 }= x_{0 }– f(x_{0})/f′(x_{0})x

_{1 }= 1.5 – (-1.25)/11.5x

_{1 }= 1.6087

In 2nd iteration:f(x

_{1}) = f(1.6087) = 2 × 1.60873 – 2 × 1.6087 – 5 = 0.1089f′(x

_{1}) = f′(1.6087) = 6 × 1.60872 – 2 = 13.52741x

_{2 }= x_{1 }– f(x_{1})/f′(x_{1})x

_{2 }= 1.6087 – 0.1089/13.52741x

_{2 }= 1.60065

In 3rd iteration:f(x

_{2}) = f(1.60065) = 2 × 1.600653 – 2 × 1.60065 – 5 = 0.00062f′(x

_{2}) = f′(1.60065) = 6 × 1.600652 – 2 = 13.37239x

_{3 }= x_{2 }– f(x_{2})/f′(x_{2})x

_{3 }= 1.60065 – 0.00062/13.37239x

_{3 }= 1.6006

The Approximate root of the equation 2x^{3 }– 2x – 5 = 0 using the Newton Raphson method is 1.6006

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