# A-level Mathematics/OCR/FP2/Numerical Methods

## The Newton-Raphson method

The best way to find a root of a function is to use the Newton-Raphson method; this uses a series of tangent lines to estimate the value of the root. The first number in the series is chosen from either one of the previous methods or semi-randomly. We can find the value of a function to a certain degree when the ${\displaystyle x_{n}=x_{n+1}}$ to the specific number of digits. This method may not work always because if f'(x) = 0 the function will be undefined at this point the function will diverge.

${\displaystyle x_{n+1}=x_{n}-{\frac {f\left(x\right)}{f'\left(x\right)}}}$

#### Example

The graph that we have above is actually the same function that Newton used to demonstrate the theory. Find the value of the root accurate to six decimal places.

We are going to use 2.05 as the first value.

${\displaystyle f'(x)=3x^{2}-2}$

${\displaystyle x_{2}=2.05-{\frac {(2.05)^{3}-2(2.05)-5}{3(2.05)^{2}-2}}=2.095710}$

Now we use this output for the next input and so on.

${\displaystyle x_{2}=2.095710-{\frac {(2.095710)^{3}-2(2.095710)-5}{3(2.095710)^{2}-2}}=2.094552}$
${\displaystyle x_{3}=2.094552-{\frac {(2.094552)^{3}-2(2.094552)-5}{3(2.094552)^{2}-2}}=2.094551}$
${\displaystyle x_{4}=2.094552-{\frac {(2.094552)^{3}-2(2.094552)-5}{3(2.094552)^{2}-2}}=2.094551}$

The root is approximately 2.094551. If we plug this into the original function we get -.00000000000004 which is very close to zero.