Solutions of Equations by Numerical Methods - Change of sign Method and Decimal Search Method.

Authors Avatar
Solutions of Equations by Numerical Methods

Change of sign Method

Decimal Search Method

f(x)=x³+2x²-3x+2

The solutions of f(x)=0 are the place where the curve cuts the x-axis.

The graph shows that there is one real root lying in the interval [-4,-3]

f(-4) = -64+32+12+2 = -18

f(-3 )= -27+18+9+2 = 2

The function f(x) is continuous therefore the change of sign form negative to positive in moving from x=-4 to x=-3 shows that there must be a root between the two values.

The decimal search method will locate more accurate values of the root systematically by taking increments of size 0.1 in the interval [-4,-3] until a change of sign occurs.

x

-4

-3.9

-3.8

-3.7

-3.6

-3.5

-3.4

-3.3

-3.2

-3.1

-3

f(x)

-18.000

-15.199

-12.592

-10.173

-7.936

-5.875

-3.984

-2.257

-0.688

0.729

2.000

There is a change of sign between -3.2 and -3.1, this show that the root is in the interval [-3.2,-3.1]

Continuing the decimal search method will increase the accuracy by narrowing down the interval further.

x

-3.20

-3.19

-3.18

-3.17

-3.16

-3.15

-3.14

-3.13

-3.12

-3.11

-3.10

f(x)

-0.688

-0.540

-0.393

-0.247

-0.103

0.039

0.180

0.320

0.457

0.594

0.729

x

-3.16

-3.159

-3.158

-3.157

-3.156

-3.155

-3.154

-3.153

-3.152

-3.151

-3.15

f(x)

-0.103

-0.089

-0.075

-0.060

-0.046

-0.032

-0.018

-0.003

0.011

0.025

0.039

x

-3.153

-3.1529

-3.1528

-3.1527

-3.1526

-3.1525

-3.1524

-3.1523

-3.1522

-3.1521

-3.152

f(x)

-0.003

-0.002

-0.001

0.001

0.002

0.004

0.005

0.007

0.008

0.009

0.011

x

-3.1528

-3.15279

-3.15278

-3.15277

-3.15276

-3.15275

-3.15274

-3.15273

-3.15272

-3.15271

-3.1527

f(x)

-0.00060

-0.00046

-0.00032

-0.00018

-0.00003

0.00011

0.00025

0.00039

0.00053

0.00068

0.00082

The decimal search method has shown that the root lies within the interval

[-3.15276,-3.15275]

The root therefore is -3.152755 with a maximum error of ± 0.000005

Failure of Decimal Search Method

f(x)=x³-2x²-3x-1/2

The graph shows that there is a real root lying in the interval [-1, 0]

f(-1) = -1-2+3-1/2 = -1/2

f( 0 ) = 0-0-0-1/2 = -1/2

The function f(x) is continuous therefore to show that there is a real root a change in sign should occur. However this is the failure of the change in sign method as the change from positive to negative is missed.

Another case where the method would fail would be when using a function which is a discontinuous curve; reciprocal. This is because there may be no roots when f(x) =0 and a change in sign may only occur due to the asymptote.

Fixed Point Estimation Method
Join now!


Newton-Raphson Method

f(x)=x³-x²-3x+1

The solutions of f(x)=0 are the place where the curve cuts the x-axis.

The graph shows that there are three real roots lying in the intervals [-2,-1], [0,1] and [2,3]

The Newton-Raphson method will locate more accurate values of the roots systematically and with a fast rate of convergence.

f(x) =x³-x²-3x+1

f '(x) =3x²-2x-3

The Iteration formula is: xn+1 = xn - f ( xn )

f '( xn )

f(x)=x³-x²-3x+1

For the interval [-2,-1] :-

X

2

3

4
...

This is a preview of the whole essay