# numerical solutions-Comparison of the three methods and Newton Raphson

Numerical Solutions of equations

1. Newton Raphson method

Equation to be solved is x³−5x−1=0  The function f(x)= x³−5x−1is shown below

There are 3 roots. I will first find the root in the interval [2, 3]

I will do the first few lines of calculation manually.

The formula to use is:  xn+1= xn –f(xn)/ f’(xn)

Therefore I must first differentiate x³−5x−1 which is 3x2-5

Using x1=3

X2:   3 – [(33-5 x 3-1)/(3 x 32 -5)] = 2.5

X3:   2.5 – [(2.53-5 x 2.5-1)/(3 x 2.52 -5)] = 2.3455

I will now work out all 3 roots using autograph until 5 significant figures are guaranteed

The 3 boxes above show how I obtained the 3 roots of the equation x³−5x−1=0

Which are -0.20164, -2.1284 and 2.3301

Below is the function f(x)= x³−5x−1 showing where I applied the Newton’s Raphsons to find each root, showing the tangent made to the curve and how it converges closer each time.

I will now illustrate one of these roots graphically, showing closely the changes in the tangent to the curve.

I will use the root in interval [2, 3]

Error bounds for interval [2, 3]

2.3301 is the root in this interval to 5 significant figures.

Therefore the error is 2.3301 ± 0.00005 I will now perform the change of sign test to confirm it is within these limits.

Lower limit is 2.33005 then f (2.33005) = -0.000098648

Upper limit is 2.33015 then f (2.33015) = 0.0010302

There is a change of sign which confirms root in interval is 2.3301 ± 0.00005

When does this method fail?

The Newton Raphson method does not always work, I will ...