Solving equations numerically

The sign - change rule

If the function f(x) is continuous for an interval a x b of its domain , if f(a) and f(b) have opposite signs, then there is at least one root of f(x) = 0 between a and b.

Consider an equation f(x) = x2 -5x + 2 = 0

-6

-4

-2

0

2

4

6

8

10

-1 0 1 2 3 4 5 6

y

x

f(0) = 2

f(1) = -2There is a sign-change

This means that there is a solution between x = 0 and x = 1.

Solving equations numerically

The sign - change ruleConsider an equation f(x) = (2x – 1)(x – 1)(2x – 3)

f(0) = -3

f(2) = 3There is a sign-change

-3

-1

1

3

-1 0 1 2x

y

It is clear from the graph that

there are three roots between

x = 0 and x = 2

ExamplesShow that a root of the equation x3 – 5x – 4 = 0 lies in the interval [2, 3].

f(2) = 23 – 5 2 – 4 = 8 – 10 – 4 = - 6 < 0

f(3) = 33 – 5 3 – 4 = 27 – 15 – 4 = 8 > 0

There is a change of sign so the root lies in the interval [2, 3]

4.32 is an approximation to a root of the equation xlnx – 2 – x = 0. Check its accuracy to 2 decimal places.

f(4.315) = 4.315 ln4.315 – 2 – 4.315 = - 0.00605 < 0

f(4.325) = 4.325 ln4.325 – 2 – 4.325 = 0.00858 > 0

Change of sign, so the root is accurate to 2 decimal places.

Decimal searchConsider f(x) = x3 – 5x – 4 ; f(2) = - 6 and f(3) = 8

f(2.5) = 2.53 – 5 2.5 – 4 = -0.875

f(2.6) = 2.63 – 5 2.6 – 4 = 0.576

f(2.55) = 2.553 – 5 2.55 – 4 = -0.169

For 1 decimal place the root is 2.5 or 2.6

x = 2.5 is ignored so the root must be x = 2.6 for one decimal place.

-10

-8

-6

-4

-2

0

2

4

6

-3 -2 -1 0 1 2 3

y

x

Use decimal search to find each root correct to two decimal places.

(i) f(x) = x + (x3 + 1) - 7

x f(x) sign

0 -6.0000 negative

1 -0.5858 negative

2 -2.0000 negative

3 0.2915 positiveA root lies between 2 and 3 and it is probably closer to 3.

2.6 -0.0900 negative

2.7 0.2479 positive

A root lies between 2.6 and 2.7 and it is probably closer to 2.6.

2.62 -0.0229 negative

2.63 0.0180 positive

A root lies between 2.62 and 2.63 and it is probably closer to 2.63.

2.625 -0.0060 negative This confirms that the root is 2.63

Use decimal search to find each root correct to two decimal places.

(ii) f(x) = x5 + x3 - 1999

x f(x) sign

2 -1959 negative

3 -1729 negative

4 -911 negative

5 1251 positiveA root lies between 4 and 5.

4.5 -62.59 negative

4.6 158 positiveA root lies between 4.5 and 4.6.

4.54 23.34 positive

4.53 1.576 positiveA root lies between 4.52 and 4.53.

4.525 -9.236 negative This confirms that the root is 4.53.

4.52 -20.00 negative

IterationConsider the equation f(x) = x2 – 5x + 2 = 0

The graph of f(x) shows that one root lies between 0 and 1 and the other root lies between 4 and 5.

-6

-4

-2

0

2

4

6

-1 0 1 2 3 4 5 6

f(x)

x

First step is rearrange x2 – 5x + 2 = 0 in the form x = g(x)

Possible rearrangements:

x2 = 5x – 2

x = (5x – 2)

5x = x2 + 2

x = (x2 + 2)/5

x2 – 5x = - 2

x(x – 5) = -2

x = 2/(5 – x)

x2 = 5x – 2

x = 5 – 2/x

Graph of rearranged equations

-6

-4

-2

0

2

4

6

-1 0 1 2 3 4 5 6

-6

-4

-2

0

2

4

6

-1 0 1 2 3 4 5 6

-6

-4

-2

0

2

4

6

-1 1 3 5

x = (5x – 2)

x = (x2 + 2)/5

-4

-2

0

2

4

6

0 1 2 3 4 5

x = 5 – 2/x

x = 2/(5-x)

Using xn+1 = g(xn)

f(x) = x2 – 5x + 2 = 0 xn+1 = (5xn – 2)

Starting value x0 = 4

x1 = (5 4 – 2) = 4.2426

x2 = 4.3833

x3 = 4.4628

x4 = 4.5071

x5 = 4.5316

x6 = 4.5451

x7 = 4.5525

x8 = 4.5566x9 = 4.5588

x10 = 4.5600

x11 = 4.5607

x12 = 4.5611

x13 = 4.5613

x14 = 4.5614

x15 = 4.5615x16 = 4.5615

4.56

root = 4.56 (2 d.p.)

f(x) = x2 – 5x + 2 = 0 xn+1 = (xn

2 + 2)/5

x0 = 4

x1 = 3.600

x2 = 2.9920

x3 = 2.1904

x4 = 1.3598

x5 = 0.7697

x6 = 0.5185

x7 = 0.4538

x8 = 0.4412

x9 = 0.4389x10 = 0.4385

x11 = 0.4385

xn+1 = 2/(5 – x)

x1 = 2.0000

x0 = 4

x2 = 1.2000

x3 = 0.6880

x4 = 0.4947

x5 = 0.4489

x6 = 0.4388

x7 = 0.4485

x8 = 0.4385

xn+1 = 5 – 2/xn

x0 = 4

x2 = 4.5556

x3 = 4.5610

x4 = 4.5615

x5 = 4.5615

x1 = 4.5000

x0 = 4.5

4.5

4.51

4.52

4.53

4.54

4.55

4.56

4.57

4.58

4.59

4.6

4.5 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59 4.6

xn+1 = 5 – 2/xn

ExampleConsider x3 – 5x – 4 = 0 , rearrange in the form xn+1 = f(xn)

x3 = 5x + 4 3 5 4x x

Let x0 = 2

x1 = 2.41012x2 = 2.52250

x3 = 2.55159

x4 = 2.55902

x5 = 2.56091

x6 = 2.56139

x7 = 2.56151

x8 = 2.56154x9 = 2.56155

x10 = 2.56155

ExampleUsing iteration method formula xn+1 = 2 + ln(xn), find a root of the equation ln(x) – x + 2 = 0, (correct to 3 s.f.) and starting with x0 = 2. x0 = 2

x1 = 2.693 147 181x2 = 2.290 710 465

x3 = 3.095 510 973

x4 = 3.129 952 989

x5 = 3.141 017 985

x6 = 3.144 546 946

x7 = 3.145 699 825

x8 = 3.146 026 848

x9 = 3.146 140 339

-1

0

1

2

3

4

0 1 2 3 4

y

x

So a root is 3.15 (3 s.f.)