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Login NowThe formula for Newton-Rapson method is

\(x_{n + 1} = x_n – \frac{f(x_n)}{f'(x_n)}\)

We have to derive this formula

**Derivation:**

Let h is an small increment in x_{n} to get next estimate of root

\(x_{n+1} – x_0 = h\)

\(x_{n+1} = x_n + h\)

We know that Taylors series can be written as

\(f(x+h) = f(x) + \frac{f'(x)h}{1!} + \frac{f”(x)h^2}{2!} + ……\)

Since h is very small, We can neglet the terms containing second and higher order terms. Then above equation can be written as:

\(f(x+h) = f(x) + \frac{f'(x)h}{1!}\)

\(f(x_n+h) = f(x_n) + \frac{f'(x_n)h}{1!}\)

\(f(x_{n+1}) = f(x_n) + \frac{f'(x_n)(x_{n+1} – x_n)}{1!}\)

If x_{n+1} is root of the given polynomail

f(x_{n+1}) = 0

\(f(x_n) + \frac{f'(x_n)(x_{n+1} – x_n)}{1!} = 0\)

Solving above equation to get the value of x_{n+1} we get

\(x_{n + 1} = x_n – \frac{f(x_n)}{f'(x_n)}\)

This is the equation of Newton Raphson formula.

**Problem Part:**

Solution:

Let f(x) = x^{4} – x – 10

f'(x) = 4x^{3} – 1

Let us assume initial guess is 1.5

x | f(x) | f'(x) | Error |

1.5 | -6.4375 | 12.5 | 0.2556 |

2.015 | 4.47 | 31.73 | 0.0896 |

1.874 | 0.4593 | 25.325 | 0.0896 |

1.855 | -0.01435 | 24.532 | 0.01024 |

Since error is less than specific limit

Root = 1.855

Therefore, Root of the equation is 1.855

**Drawback of Newton-Rapson Method:**

- It’s convergence is not guaranteed. So, sometimes, for given equation and for given guess we may not get solution.
- Division by zero problem can occur.
- Root jumping might take place thereby not getting intended solution.
- Inflection point issue might occur.
- Symbolic derivative is required.
- In case of multiple roots, this method converges slowly.
- Near local maxima and local minima, due to oscillation, its convergence is slow.

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