## - 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.

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, $$α=1.15$$.

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 $$x_4$$.