Derive the formula for Newton Raphson method. Solve the equation x4 – x – 10 = 0 using Newton Raphson method assuming error precision is 0.01. Discuss drawback for this method.

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The formula for Newton-Rapson method is

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

We have to derive this formula


Let h is an small increment in xn 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 xn+1 is root of the given polynomail

f(xn+1) = 0

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

Solving above equation to get the value of xn+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:


Let f(x) = x4 – x – 10

f'(x) = 4x3 – 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:

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


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