- Newton Raphson

The gradient of the tangent at the point \((x_n,f(x_n))\) is the derivative of \(f(x)\) i.e. \(f'(x_n)\). Therefore,

\[f'(x_n)={f(x_n)-0 \over x_n-x_{n+1}}\]
Solving for \(x_{n+1}\)
\[f'(x_n)[x_n-x_{n+1}]=f(x_n)\]
\[x_n-x_{n+1}={f(x_n)\over f'(x_n)}\]
\[x_{n+1}={x_n-f(x_n)\over f'(x_n)}\]
This is the iterative formula known as the Newton – Raphson formula.

VIDEO EXPLANATION


LESSON 1
The equation \(x^3-x^2+4x-900=0\) has exactly one real root, \(α\). Taking \(x_1=10\) as the first approximation to α, use the Newton – Raphson method to find a second approximation, \(x_2\), to \(α\). Give your answer to four significant figures.
SOLUTION
\[f(x)=x^3-x^2+4x-900\]
Determine \(f'(x)\)
\[f'(x)=3x^2-2x+4\]
Let \(x_1=10\)
Use the Newton – Raphson formula
\[x_2={x_1-f(x_1)\over f'(x_1)}\]
\[x_2={10-10^3-10^2+4(10)-900 \over 3(10)^2-2(10)+4}\]
\[x_2={10-40\over 284}\]
\[x_2=9.859\]

LESSON 2
The equation \(x^3+5x^2+3x-13=0\) has exactly one real root \(α\).
(i) Show that \(α\) lies in the interval \(1<α<2\).
(ii) Using the Newton – Raphson method with initial estimate \(x_1=1.5\) to estimate the root of the equation \(2x^3+5x^2+3x-13=0\) in the interval [1,2], correct to 2 decimal places.
SOLUTION
(i) Let \(f(x)=2x^3+5x^2+3x-13\)
\[f(1)=2(1)^3+5(1)^2+3(1)-13=-3\]
\[f(2)=2^3+5(2)^2+3(2)-13=29\]
Since \(f(1)×f(2)<0\), by the Intermediate Value Theorem there is a root in the interval [1,2]
(ii)
\[f(x)=2x^3+5x^2+3x-13\]
Determine \(f'(x)\)
\[f'(x)=6x^2+10x+3\]
Let \(x_1=1.5\) and use the Newton – Raphson formula
\[x_2={1.5-2(1.5)^3+5(1.5)^2+3(1.5)-13\over 6(1.5)^2+10(1.5)+3}\]
\[x_2={1.5-19\over 63}=1.1984\]
Substitute the value of \(x_2\) into the Newton – Raphson formula to determine \(x_3\).
\[x_3={1.1984-2(1.1984)^3+5(1.1984)^2+3(1.1984)-13\over 6(1.1984)^2+10(1.1984)+3}\]
\[x_3=1.1984-0.0516\]
\[x_3=1.1467\]
Substitute the value of \(x_3\) into the Newton – Raphson formula to determine \(x_4\).
\[x_4={1.1467-2(1.1467)^3+5(1.1467)^2+3(1.1467)-13\over 6(1.1467)^2+10(1.1467)+3}\]
\[x_4=1.1984-0.001357\]
\[x_4=1.1453\]
Since \(x_3\) and \(x_4\) correct to 2 decimal places are both equal to 1.15, \(α=1.15\).

VIDEO SOLUTION


USING THE CALCULATOR
The above calculations can be easily achieved through the use of the ANS function on the calculator.
On the calculator
  • Type the value of \(x_1\) into the calculator and press ANS so that the calculator stores this value.
  • To determine \(x_2\), type
\[{ANS-2(ANS)^3+5(ANS)^2+3(ANS)-13\over 6(ANS)^2+10(ANS)+3}\]
  • Press the equal sign to get the value of \(x_2\).
  • Press the equal sign to get the value of \(x_3\).
  • Press the equal sign to get the value of \(x_4\).

VIDEO EXPLANATION