Solution of Transcendental & Algebraic Equations

Dr. Sandeep Malhotra

Depar tment of Mathematics & Humanit ies,

Inst i tute of Technology,

Nirma Universi ty.

B. Tech. in Electrical Engineering (Semester III)

Electrical Engineering Problem

Find the resistance, R, of a circuit such that the charge reaches q atspecified time t,

(1.1)t

2-Rt 2L

0

1 R qf(R) = e cos - - = 0

LC 2L q

Where, L = inductance,

C = capacitance,

q0 = initial charge

Force on Falling Parachutist

By Newton second law of motion, which states that the time rate of change of momentum of a body is equal to the resultant force acting on it.

The mathematical expression, or model, of the second law is the well-known equation

F = ma (1.2)

⟹ 𝑭 = 𝒎𝒅𝒗

𝒅𝒕(1.3)

The force acting on the body : F = FU+ FD

Where, FD= mg & FU = -cv

c is a proportionality constant called the drag coefficient (kg/s).

FU

FD

Therefor, by Eq. (1.3)

This is a 1st Order Linear ODE, solving which we get,

𝒗 𝒕 =𝒈𝒎

𝒄𝟏 − 𝒆−

𝒄

𝒎𝒕

(1.5)

If the parameters are known, Eq. (1.5) can be used to predict theparachutist’s velocity as a function of time. Such computations can beperformed directly because v is expressed explicitly as a function of time. Thatis, it is isolated on one side of the equal sign.

mg - c𝒗 = 𝒎𝒅𝒗

𝒅𝒕

⟹𝒅𝒗

𝒅𝒕= 𝒈 −

𝒄

𝒎𝒗 (1.4)

However, suppose we had to determine the drag coefficient for a parachutistof a given mass to attain a prescribed velocity in a set time period.

Although Eq. (1.5) provides a mathematical representation of theinterrelationship among the model variables and parameters, it cannot besolved explicitly for the drag coefficient.

There is no way to rearrange the equation so that c is isolated on one side ofthe equal sign. In such cases, c is said to be implicit.

This represents a real dilemma, because many engineering design problemsinvolve specifying the properties or composition of a system (as representedby its parameters) to ensure that it performs in a desired manner (asrepresented by its variables).

Thus, these problems often require the determination of implicit parameters.

To solve the problem, it is conventional to re-express Eq. (1.5).

This is done by subtracting the dependent variable v from both sides ofthe equation to give,

The value of c that makes f(c) = 0 is, therefore, the root of the equation.

The solution to the dilemma is provided by numerical methods for roots ofalgebraic or transcendental equations.

𝒇 𝒄 =𝒈𝒎

𝒄𝟏 − 𝒆−

𝒄

𝒎𝒕 − 𝒗 (1.6)

Algebraic & Transcendental Equations By definition, a function given by y = 𝒇 𝒙 = 0 is algebraic if it can beexpressed in the form

𝒇 𝒙 = 𝒂𝒏𝒙𝒏 + 𝒂𝒏−𝟏𝒙

𝒏−𝟏 +⋯+ 𝒂𝟏𝒙 + 𝒂𝟎 = 𝟎.

Some specific examples are,

𝒇 𝒙 = 𝟏 − 𝟐. 𝟑𝟕𝒙 + 𝟕. 𝟓𝒙𝟐 = 𝟎 & 𝒇 𝒙 = 𝟕𝒙𝟔 − 𝒙𝟑 + 𝟓𝒙𝟐 = 𝟎

A transcendental function is one that is non algebraic. These includetrigonometric, exponential, logarithmic, and other, less familiar, functions.

Examples are,

𝒇 𝒙 = 𝐥𝐧 𝒙𝟐 − 𝟏 = 𝟎 & 𝒇 𝒙 = 𝒆−𝟎.𝟐𝒙 𝐬𝐢𝐧 𝟑𝒙 − 𝟎. 𝟓 = 𝟎

by Lale Yurttas, Texas A&M University PART 2 8

Why new methods to find Roots of Equations?

• BUT

a

acbbxcbxax

2

40

22

?0sin

?02345

xxx

xfexdxcxbxax

Root finding Methods Essentially finding a root of a function, that is, a zero of the function.

The goal is to solve f(x) = 0, for the function f(x).

The values of x which make f(x) = 0 are the roots of the equation.

f(a)=0 f(b)=0

a, b are roots of the function f(x)

X

f(x)

Y= f(x)

Root finding Methods There are many methods for solving non-linear equations. The methods, which willbe highlighted have the following properties: The function f(x) is expected to be continuous. If not the method may fail.

The use of the algorithm requires that the solution be bounded.

Once the solution is bounded, it is refined to specified tolerance.

Such methods are:

1. Graphical Method

2. Interval Halving (Bisection method)

3. Regula Falsi (False position)

4. Newton’s method

5. Secant method

6. Fixed point iteration

Study Objectives

You should have sufficient information to successfully approach a

wide variety of engineering problems dealing with roots of

equations.

You should have

mastered the techniques,

have learned to assess their reliability,

and be capable of choosing the best method (or methods) for any

particular problem.

Important Instructions

You must have Scientific Calculator in each class.

Fix your Calculator digits to 6 decimals.

Put it in Radian mode.

Don’t cut the digits specially when you are usingthe Secant Method.

Nonlinear Equation

Solvers

Bracketing

Bisection

False Position

(Regula-Falsi)

Graphical Open Methods

Newton Raphson

Secant

All Iterative

Fixed Point

GRAPHICAL METHOD

Graphical MethodA simple method for obtaining an estimate of the root of the equation f(x) = 0is to make a plot of the function and observe where it crosses the x axis.

This point, which represents the x value for which f(x) = 0, provides a roughapproximation of the root.

Illustration: Use the graphical approach to determine the drag coefficient cneeded for a parachutist of mass m = 68.1 kg to have a velocity of 40 m /safter free-falling for time t = 10 s.

Solution: This problem can be solved by determining the root of Eq. (1.6) usingthe parameters t = 10, g = 9.8, v = 40, and m = 68.1:

⟹ 𝒇 𝒄 =𝟗.𝟖(𝟔𝟖.𝟏)

𝒄𝟏 − 𝒆−(

𝒄

𝟔𝟖.𝟏)𝟏𝟎 − 40𝒇 𝒄 =

𝒈𝒎

𝒄𝟏 − 𝒆−

𝒄𝒎𝒕 − 𝒗

Various values of c can be substituted into the right-hand side of this equation

to compute, These points are plotted in Fig. below,

The resulting curve crosses the c axis between 12

and 16. Visual inspection of the plot provides a

rough estimate of the root of 14.75. The validity

of the graphical estimate can be checked by

substituting it into Eq. (1.7) to yield,

⟹ 𝒇 𝒄 =𝟔𝟔𝟕.𝟑𝟖

𝒄𝟏 − 𝒆−𝟎.𝟏𝟒𝟔𝟖𝟒𝟑𝒄 −40 (1.7)

𝒇 𝟏𝟒. 𝟕𝟓 =𝟔𝟔𝟕.𝟑𝟖

𝟏𝟒.𝟕𝟓𝟏 − 𝒆−𝟎.𝟏𝟒𝟔𝟖𝟒𝟑(𝟏𝟒.𝟕𝟓) − 40

= 0.059

40

20

0

-10

4 8 12 16 20c

f(c)

Root

c f(c)

4 34.115

8 17.653

12 6.067

16 -2.269

20 -8.401

Example 1: Determine the real root of 𝒇 𝒙 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 + 𝟔𝒙− 𝟐 graphically.

Solution: Lets prepare a table and graph of values for x and f(x),

𝑥 = 0.4 ⟹ 𝑓 𝑥 = −0.08

𝑥 = 0.42 ⟹ 𝑓 𝑥 = 0.008 ≈ 0.

x f(x)

0 -2

1 4

2 30

-1 -18

-1 1 2 x

f(x)

Root

40

20

0

-20

The resulting curve crosses the x axis between 0

and 1. Visual inspection of the plot provides a

rough estimate of the root of 0.4. The validity of

the graphical estimate can be checked by

substituting it into original Eq. to yield,

Ans.: Approximate root of the equation is x = 0.42.

Pitfalls of Graphical Method Graphical technique is of limited practical value because it is not precise.

However, graphical methods can be utilized to obtain rough estimates of roots. Theseestimates can be employed as starting guesses for numerical methods discussed in the nexttopic.

From providing rough estimates of the root, graphical interpretations are importanttools for understanding the properties of the functions and anticipating the pitfalls ofthe numerical methods.

Intermediate Value property: The function changes sign on opposite sides of the root.If one root of a real and continuous function, f(x) = 0, is bounded by values x = a, x = bthen f(a) . f(b) < 0.

It is the simplest root-finding step which requires previous knowledge of two initialguesses, a and b, so that f(a) and f(b) have opposite signs. These guesses must “bracket”or be on either side of the root.

Bracketing Methods

In bracketing methods, the method starts with an intervalthat contains the root and a procedure is used to obtain asmaller interval containing the root.

Such methods are always convergent.

Examples of bracketing methods:

◦Bisection method

◦False position method

Bisection Method

Bisection Method Assumptions:

Given an interval [a, b]

f(x) is continuous on [a, b]

f(a) and f(b) have opposite signs.

These assumptions ensure the existence of at least one zero in the interval [a, b] and thebisection method can be used to obtain a smaller interval that contains the zero.

For that we perform the following steps:

1. Compute the mid point c = (a + b) / 2

2. Evaluate f(c)

3. If f(a) f(c) < 0 then new interval [a, c]

If f(a) f(c) > 0 then new interval [c, b]

4. Repeat the procedure until we get convergence.

a

b

f(a)

f(b)

c

a C1 C2

b

Example 2: Determine the real root of 𝒇 𝒙 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 + 𝟔𝒙 − 𝟐 = 𝟎 using Bisection Method.

Solution: Lets find the interval first,

𝑥 = 0 ⟹ 𝑓 0 = −2𝑥 = 1 ⟹ 𝑓 1 = 4∴ 𝑥 ∈ [0, 1]

Lets start the Bisection iterations now,

𝑥1 =0+1

2= 0.5, 𝑓 𝑥1 = 𝑓 0.5 = 0.375 > 0 ∴ 𝑥 ∈ [0, 0.5]

𝑥2 =0 + 0.5

2= 0.25, 𝑓 𝑥2 = 𝑓 0.25 = -0.7344 < 0 ∴ 𝑥 ∈ [0.25, 0.5]

𝑥3 =0.25 + 0.5

2= 0.375, 𝑓 𝑥3 = 𝑓 0.375 =-0.1895<0 ∴ 𝑥 ∈ [0.375, 0.5]

< 0> 0

Table Showing The Iterations

a = 0, f(a) < 0 b = 1, f(b) > 0 x1 = 0.5 f(x1) = 0.375 > 0

a = 0 x1 = 0.5 x2 = 0.25 f(x2) = -0.7344 < 0

x2 = 0.25 x1 = 0.5 x3 = 0.375 f(x3) = 0.1895 > 0

x3 = 0.375 X1 = 0.5 x4 = f(x4) =

𝑥4 =0.375 + 0.5

2= 0.4375, 𝑓 𝑥4 = 𝑓 0.4375 = 0.0867 > 0 ∴ 𝑥 ∈ [0.375, 0.4375]

𝑥5 =0.375 + 0.4375

2= 0.4063, 𝑓 𝑥5 = 𝑓 0.4063 = -0.0522 < 0 ∴ 𝑥 ∈ [0.4063, 0.4375]

𝑥6 =0.4063 + 0.4375

2= 0.4219,𝑓 𝑥6 = 𝑓 0.4219 = 0.0169 > 0 ∴ 𝑥 ∈ [0.4063, 0.4219]

𝑥7 =0.4063 + 0.4219

2= 0.4141,𝑓 𝑥7 = 𝑓 0.4141 = -0.0177 < 0 ∴ 𝑥 ∈ [0.4141, 0.4219]

𝑥8 =0.4141 + 0.4219

2= 0.418, 𝑓 𝑥8 = 𝑓 0.418 = -0.00044 < 0 ∴ 𝑥 ∈ [0.418, 0.4219]

𝑥9 =0.418 + 0.4219

2= 0.42, 𝑓 𝑥9 = 𝑓 0.42 = 0.0084 > 0 ∴ 𝑥 ∈ [0.42, 0.4219]

|0.42 – 0.4219| = 0.0019 ≈ 0

∴ 𝒙 = 𝟎. 𝟒𝟐 is one of the approximate real root of the given equation

correct up to 2 decimal places.

Example 3: Determine the negative real root of 𝒇 𝒙 = 𝒙𝟑 + 𝟐𝟏𝒙 + 35

using Bisection Method correct up to 2 decimal places.

Solution: Lets find the interval first, 𝑥 = 0 ⟹ 𝑓 0 = 35

𝑥 = −1 ⟹ 𝑓 −1 =13

∴ 𝑥 ∈ [−2,−1]

Lets start the Bisection iterations now,

𝑥1 =−2−1

2= −1.5, 𝑓 𝑥1 = 𝑓 −1.5 = 0.125 > 0 ∴ 𝑥 ∈ [−2,−1.5]

𝑥2 =−2 − 1.5

2= −1.75, 𝑓 𝑥2 = 𝑓 −1.75 = -7.12 < 0 ∴ 𝑥 ∈ [−1.75,−1.5]

𝑥3 =−1.75 − 1.5

2= −1.625, 𝑓 𝑥3 = 𝑓 −1.625 = -3.42 < 0

∴ 𝑥 ∈ [−1.625, −1.5]

> 0

> 0

𝑥 = −2 ⟹ 𝑓 −2 = -15 < 0

𝑥4 =−1.625 − 1.5

2= −1.563, 𝑓 𝑥4 = 𝑓 −1.563 = -1.641 < 0 ∴ 𝑥 ∈ [−1.563,−1.5]

𝑥5 =−1.563 − 1.5

2= −1.532, 𝑓 𝑥5 = 𝑓 −1.532 = -0.768 < 0 ∴ 𝑥 ∈ [−1.532,−1.5]

𝑥6 =−1.532 − 1.5

2= −1.516, 𝑓 𝑥6 = 𝑓 −1.516 = -0.32 < 0 ∴ 𝑥 ∈ [−1.516,−1.5]

𝑥7 =−1.516 − 1.5

2= −1.508, 𝑓 𝑥7 = 𝑓 −1.508 = -0.097 < 0 ∴ 𝑥 ∈ [−1.508,−1.5]

𝑥8 =−1.508 − 1.5

2= −1.504, 𝑓 𝑥8 = 𝑓 −1.504 = 0.014 > 0 ∴ 𝑥 ∈ [−1.508,−1.504]

|(-1.508) – (-1.504)| = 0.004 ≈ 0

∴ 𝒙 = −𝟏.504 is one of the approximate negative real root of the given equation

correct up to 2 decimal places.

Example 4: Determine the nontrivial root of 𝒔𝒊𝒏 𝒙 = 𝒙𝟑 using Bisection Method correct up to 2 decimal places.

Solution: Lets find the interval for 𝑓 𝑥 = 𝑠𝑖𝑛𝑥 − 𝑥3 = 0,

𝑥 = 0.5 ⟹ 𝑓 0.5 = 0.3544

∴ 𝑥 ∈ [0.5, 1]

Lets start the Bisection iterations now,

𝑥1 =0.5+1

2= 0.75, 𝑓 𝑥1 = 𝑓 0.75 = 0.26 > 0 ∴ 𝑥 ∈ [0.75, 1]

𝑥2 =0.75 + 1

2= 0.875, 𝑓 𝑥2 = 𝑓 0.875 = 0.098 > 0 ∴ 𝑥 ∈ [0.875, 1]

𝑥3 =0.875 + 1

2= 0.9375, 𝑓 𝑥3 = 𝑓 0.9375 = -0.018 < 0

∴ 𝑥 ∈ [0.875, 0.9375]

> 0

𝑥 = 1 ⟹ 𝑓 1 = -0.159 < 0

𝑥4 =0.875 + 0.9375

2= 0.9063, 𝑓 𝑥4 = 𝑓 0.9063 = 0.0428 > 0 ∴ 𝑥 ∈ [0.9063, 0.9375]

𝑥5 =0.9063 + 0.9375

2= 0.9219, 𝑓 𝑥5 = 𝑓 0.9219 = 0.013 > 0 ∴ 𝑥 ∈ [0.9219, 0.9375]

𝑥6 =0.9219 + 0.9375

2= 0.9297,𝑓 𝑥6 = 𝑓 0.9297 = -0.002 < 0 ∴ 𝑥 ∈ [0.9219, 0.9297]

|0.9219 - 0.9297| = 0.0078 ≈ 0

∴ 𝒙 = 𝟎. 𝟗𝟐 is one of the approximate real root of the given equation

correct up to 2 decimal places.

Pitfalls of Bisection MethodPros

Easy

Always find root

Number of iterationsrequired to attain anabsolute error can becomputed a priori as thesize of the intervalcontaining the zero isreduced by 50% aftereach iteration.

Cons

Slow

Only find one root at atime.

Not applicable whenthere are Multiple roots.

Good intermediateapproximations may bediscarded.

i 1 imax maxE 0.5 E

False Position Method(Regula Falsi)

Method of False Position

This technique is similar to the bisection

method except that the next iteration

is taken as the line of interception

between the pair of x-values and the

x-axis rather than at the midpoint.

(a, f(a))

(b, f(b))

By two point line formula,

𝒚 − 𝒚𝟏𝒚𝟏 − 𝒚𝟐

=𝒙 − 𝒙𝟏𝒙𝟏 − 𝒙𝟐

⟹𝒇(𝒙) − 𝒇(𝒂)

𝒇(𝒂) − 𝒇(𝒃)=𝒙 − 𝒂

𝒂 − 𝒃

⟹−𝒇(𝒂)

𝒇(𝒂) − 𝒇(𝒃)=𝒙𝟏 − 𝒂

𝒂 − 𝒃𝒇𝒐𝒓 𝒙 = 𝒙𝟏, 𝒇 𝒙 = 𝟎 ⟹ 𝒙𝟏 = 𝒂 −

(𝒂 − 𝒃)𝒇(𝒂)

𝒇(𝒂) − 𝒇(𝒃)

⟹ 𝒙𝟏 =−𝒂𝒇(𝒃) + 𝒃𝒇(𝒂)

𝒇(𝒂) − 𝒇(𝒃)⟹ 𝒙𝟏 =

𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)

𝒇(𝒃) − 𝒇(𝒂)Iterative formula for Method

of False Position

X

f(x)

Roota

b

𝒙𝟏 𝒙𝟐

Method of False PositionAssumptions:

Given an interval [a, b]

f(x) is continuous on [a, b]

f(a) and f(b) have opposite signs.

These assumptions ensure the existence of at least one zero in the interval [a, b] and thefalse position method can be used to obtain a smaller interval that contains the zero.

For that we perform the following steps:

1. Compute the in between point on x-axis,

2. Evaluate f(𝒙𝟏)

3. If f(a) f(𝒙𝟏) < 0 then new interval [a, 𝒙𝟏]

If f(a) f(𝒙𝟏) > 0 then new interval [𝒙𝟏, b]

4. Repeat the procedure until 𝒇(𝒙𝒊) < tolerance value.

𝒙𝟏 =𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)

𝒇(𝒃) − 𝒇(𝒂)

Solution: Lets find the interval first,

𝑥 = 0 ⟹ 𝑓 0 = −2 < 0𝑥 = 1 ⟹ 𝑓 1 = 4 > 0∴ 𝑥 ∈ [0, 1]

Lets start the iterations now,

Example 5: Determine the real root of

𝒇 𝒙 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 + 𝟔𝒙 − 𝟐 = 𝟎 using Method of False Position.

𝒙𝟏 =𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)

𝒇(𝒃) − 𝒇(𝒂)=𝟎𝒇 𝟏 − 𝟏𝒇(𝟎)

𝒇(𝟏) − 𝒇(𝟎)=𝟎 − 𝟏(−𝟐)

𝟒 − (−𝟐)=𝟐

𝟔= 𝟎. 𝟑𝟑𝟑

𝒇 𝒙𝟏 = 𝒇 𝟎. 𝟑𝟑𝟑 = -0.372 < 0 ∴ 𝒙 ∈ [𝟎. 𝟑𝟑𝟑, 𝟏]

Table Showing The Iterations

a = 0, f(a) < 0 b = 1, f(b) > 0 x1 = 0.333 f(x1) = -0.372 < 0

x1 = 0.333 b = 1 x2 = 0.3898 f(x2) = -0.125 < 0

x2 = 0.3898 b = 1 x3 = 0.4083 f(x3) = -0.043 < 0

x3 = 0.4083 b = 1 x4 = f(x4) =

𝒙𝟐 =𝒙𝟏𝒇 𝒃 − 𝒃𝒇(𝒙𝟏)

𝒇(𝒃) − 𝒇(𝒙𝟏)=𝟎. 𝟑𝟑𝟑𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟑𝟑𝟑)

𝒇(𝟏) − 𝒇(𝟎. 𝟑𝟑𝟑)=𝟎. 𝟑𝟑𝟑(𝟒) − 𝟏(−𝟎. 𝟑𝟕𝟐)

𝟒 − (−𝟎. 𝟑𝟕𝟐)= 𝟎. 𝟑𝟖𝟗𝟖

𝒇 𝒙𝟐 = 𝒇 𝟎. 𝟑𝟖𝟗𝟖 = -0.125 < 0 ∴ 𝒙 ∈ [𝟎. 𝟑𝟖𝟗𝟖, 𝟏]

𝒙𝟑 =𝒙𝟐𝒇 𝒃 − 𝒃𝒇(𝒙𝟐)

𝒇(𝒃) − 𝒇(𝒙𝟐)=𝟎. 𝟑𝟖𝟗𝟖𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟑𝟖𝟗𝟖)

𝒇(𝟏) − 𝒇(𝟎. 𝟑𝟖𝟗𝟖)=𝟎. 𝟑𝟖𝟗𝟖(𝟒) − 𝟏(−𝟎. 𝟏𝟐𝟓)

𝟒 − (−𝟎. 𝟏𝟐𝟓)= 𝟎.4083

𝒇 𝒙𝟑 = 𝒇 𝟎. 𝟒𝟎𝟖𝟑 = -0.043 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟎𝟖𝟑, 𝟏]

𝒙𝟒 =𝒙𝟑𝒇 𝒃 − 𝒃𝒇(𝒙𝟑)

𝒇(𝒃) − 𝒇(𝒙𝟑)=𝟎. 𝟒𝟎𝟖𝟑𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟎𝟖𝟑)

𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟎𝟖𝟑)=𝟎. 𝟒𝟎𝟖𝟑(𝟒) − 𝟏(−𝟎. 𝟎𝟒𝟑)

𝟒 − (−𝟎. 𝟎𝟒𝟑)= 𝟎.415

𝒇 𝒙𝟒 = 𝒇 𝟎. 𝟒𝟏𝟓 = -0.014 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟓, 𝟏]

𝒙𝟓 =𝒙𝟒𝒇 𝒃 − 𝒃𝒇(𝒙𝟒)

𝒇(𝒃) − 𝒇(𝒙𝟒)=𝟎. 𝟒𝟏𝟓𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟏𝟓)

𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟏𝟓)=𝟎. 𝟒𝟏𝟓(𝟒) − 𝟏(−𝟎. 𝟎𝟏𝟒)

𝟒 − (−𝟎. 𝟎𝟏𝟒)= 𝟎.417

𝒇 𝒙𝟓 = 𝒇 𝟎. 𝟒𝟏𝟕 = -0.005 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟕, 𝟏]

𝒙𝟔 =𝒙𝟓𝒇 𝒃 − 𝒃𝒇(𝒙𝟓)

𝒇(𝒃) − 𝒇(𝒙𝟓)=𝟎. 𝟒𝟏𝟕𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟏𝟕)

𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟏𝟕)=𝟎. 𝟒𝟏𝟕(𝟒) − 𝟏(−𝟎. 𝟎𝟎𝟓)

𝟒 − (−𝟎. 𝟎𝟎𝟓)= 𝟎.4177

𝒇 𝒙𝟔 = 𝒇 𝟎. 𝟒𝟏𝟕𝟕 = -0.0018 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟕𝟕, 𝟏]

𝒙𝟕 =𝒙𝟔𝒇 𝒃 − 𝒃𝒇(𝒙𝟔)

𝒇(𝒃) − 𝒇(𝒙𝟔)=𝟎. 𝟒𝟏𝟕𝟕𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟏𝟕𝟕)

𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟏𝟕𝟕)=𝟎. 𝟒𝟏𝟕𝟕(𝟒) − 𝟏(−𝟎. 𝟎𝟎𝟏𝟖)

𝟒 − (−𝟎. 𝟎𝟎𝟏𝟖)= 𝟎.4179

𝒇 𝒙𝟕 = 𝒇 𝟎. 𝟒𝟏𝟕𝟗 = -0.0009 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟕𝟗, 𝟏]

∴ 𝒙 = 𝟎. 𝟒𝟏𝟕𝟗 is one of the approximate real root of the given equation

correct up to 3 decimal places.

|f(0.4179)| = 0.0009 ≈ 0

Example 6: Determine the real root of 𝒍𝒏 𝒙𝟒 = 𝟎. 𝟕 using Regula – Falsi

Method correct up to three decimal places.

Solution: Lets find the interval for,

𝑥 = 1 ⟹ 𝑓 1 = −0.7𝑥 = 2 ⟹ 𝑓 2 = 2.073∴ 𝑥 ∈ [1, 2]

Lets start the iterations now,

𝒇 𝒙 = 𝒍𝒏 𝒙𝟒 − 𝟎. 𝟕

𝒙𝟏 =𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)

𝒇(𝒃) − 𝒇(𝒂)=𝟏𝒇 𝟐 − 𝟐𝒇(𝟏)

𝒇(𝟐) − 𝒇(𝟏)=𝟐. 𝟎𝟕𝟑 − 𝟐(−𝟎. 𝟕)

𝟐. 𝟎𝟕𝟑 − (−𝟎. 𝟕)= 𝟏. 𝟐𝟓𝟐

𝒇 𝒙𝟏 = 𝒇 𝟏. 𝟐𝟓𝟐 = 0.199 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟐𝟓𝟐]

< 0> 0

Table Showing The Iterations

a = 1, f(a) < 0 b = 2, f(b) > 0 x1 = 1.252 f(x1) = 0.199 > 0

a = 1 x1 = 1.252 x2 = 1.196 f(x2) = 0.016 > 0

a = 1 x2 = 1.196 x3 = f(x3) =

𝒙𝟐 =𝒂𝒇 𝒙𝟏 − 𝒙𝟏𝒇(𝒂)

𝒇(𝒙𝟏) − 𝒇(𝒂)=𝟏𝒇 𝟏. 𝟐𝟓𝟐 − 𝟏. 𝟐𝟓𝟐𝒇(𝟏)

𝒇(𝟏. 𝟐𝟓𝟐) − 𝒇(𝟏)=𝟎. 𝟏𝟗𝟗 − 𝟏. 𝟐𝟓𝟐(−𝟎. 𝟕)

𝟎. 𝟏𝟗𝟗 − (−𝟎. 𝟕)= 𝟏.196

𝒇 𝒙𝟐 = 𝒇 𝟏. 𝟏𝟗𝟔 = 0.016 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟔]

𝒙𝟑 =𝒂𝒇 𝒙𝟐 − 𝒙𝟐𝒇(𝒂)

𝒇(𝒙𝟐) − 𝒇(𝒂)=𝟏𝒇 𝟏. 𝟏𝟗𝟔 − 𝟏. 𝟏𝟗𝟔𝒇(𝟏)

𝒇(𝟏. 𝟏𝟗𝟔) − 𝒇(𝟏)=𝟎. 𝟎𝟏𝟔 − 𝟏. 𝟏𝟗𝟔(−𝟎. 𝟕)

𝟎. 𝟎𝟏𝟔 − (−𝟎. 𝟕)= 𝟏.192

𝒇 𝒙𝟑 = 𝒇 𝟏. 𝟏𝟗𝟐 = 0.0025 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟐]

𝒙𝟒 =𝒂𝒇 𝒙𝟑 − 𝒙𝟑𝒇(𝒂)

𝒇(𝒙𝟑) − 𝒇(𝒂)=𝟏𝒇 𝟏. 𝟏𝟗𝟐 − 𝟏. 𝟏𝟗𝟐𝒇(𝟏)

𝒇(𝟏. 𝟏𝟗𝟐) − 𝒇(𝟏)=𝟎. 𝟎𝟎𝟐𝟓 − 𝟏. 𝟏𝟗𝟐(−𝟎. 𝟕)

𝟎. 𝟎𝟎𝟐𝟓 − (−𝟎. 𝟕)= 𝟏.1913

𝒇 𝒙𝟒 = 𝒇 𝟏. 𝟏𝟗𝟏𝟑 = 0.00002 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟏𝟑]

𝒙𝟓 =𝒂𝒇 𝒙𝟒 − 𝒙𝟒𝒇(𝒂)

𝒇(𝒙𝟒) − 𝒇(𝒂)=𝟏𝒇 𝟏. 𝟏𝟗𝟏𝟑 − 𝟏. 𝟏𝟗𝟏𝟑𝒇(𝟏)

𝒇(𝟏. 𝟏𝟗𝟐) − 𝒇(𝟏)=𝟎. 𝟎𝟎𝟎𝟎𝟐 − 𝟏. 𝟏𝟗𝟏𝟑(−𝟎. 𝟕)

𝟎. 𝟎𝟎𝟎𝟎𝟐 − (−𝟎. 𝟕)= 𝟏.19129

𝒇 𝒙𝟓 = 𝒇 𝟏. 𝟏𝟗𝟏𝟐𝟗 = 0.00001 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟏𝟐𝟗]

∴ 𝒙 = 𝟏. 𝟏𝟗𝟏𝟐𝟗 is the approximate real root of the given equation correct up to 3 decimal places.

Example 7: Determine the real root of 𝒙𝟏𝟎= 1.1 using Method of False

Position & Bisection Method correct up to three decimal places.

Solution: Lets find the interval for, 𝒇 𝒙 = 𝒙𝟏𝟎 − 1.1

𝑥 = 1 ⟹ 𝑓 1 = −0.1𝑥 = 2 ⟹ 𝑓 2 = 1022.9

< 0> 0

∴ 𝑥 ∈ [1, 2]

Part 1. Iterations for Method of False Position

𝒙𝟏 =𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)

𝒇(𝒃) − 𝒇(𝒂)=𝟏𝒇 𝟐 − 𝟐𝒇(𝟏)

𝒇(𝟐) − 𝒇(𝟏)=𝟏𝟎𝟐𝟐. 𝟗 − 𝟐(−𝟎. 𝟏)

𝟏𝟎𝟐𝟐. 𝟗 − (−𝟎. 𝟏)= 𝟏.0001

𝒇 𝒙𝟏 = 𝒇 𝟏. 𝟎𝟎𝟎𝟏 = -0.099 < 0 ∴ 𝒙 ∈ [𝟏. 𝟎𝟎𝟎𝟏, 𝟐]

𝒙𝟐 =𝒙𝟏𝒇 𝒃 − 𝒃𝒇(𝒙𝟏)

𝒇(𝒃) − 𝒇(𝒙𝟏)=𝟏. 𝟎𝟎𝟎𝟏𝒇 𝟐 − 𝟐𝒇(𝟏. 𝟎𝟎𝟎𝟏)

𝒇(𝟐) − 𝒇(𝟏. 𝟎𝟎𝟎𝟏)=(𝟏. 𝟎𝟎𝟎𝟏)(𝟏𝟎𝟐𝟐. 𝟗) − 𝟐(−𝟎. 𝟎𝟗𝟗)

𝟏𝟎𝟐𝟐. 𝟗 − (−𝟎. 𝟎𝟗𝟗)

= 𝟏.0002𝒇 𝒙𝟐 = 𝒇 𝟏. 𝟎𝟎𝟎𝟐 = -0.098 < 0 ∴ 𝒙 ∈ [𝟏. 𝟎𝟎𝟎𝟐, 𝟐]

𝒙𝟑 =𝒙𝟐𝒇 𝒃 − 𝒃𝒇(𝒙𝟐)

𝒇(𝒃) − 𝒇(𝒙𝟐)=𝟏. 𝟎𝟎𝟎𝟐𝒇 𝟐 − 𝟐𝒇(𝟏. 𝟎𝟎𝟎𝟐)

𝒇(𝟐) − 𝒇(𝟏. 𝟎𝟎𝟎𝟐)=(𝟏. 𝟎𝟎𝟎𝟐)(𝟏𝟎𝟐𝟐. 𝟗) − 𝟐(−𝟎. 𝟎𝟗𝟖)

𝟏𝟎𝟐𝟐. 𝟗 − (−𝟎. 𝟎𝟗𝟖)

= 𝟏.0003𝒇 𝒙𝟑 = 𝒇 𝟏. 𝟎𝟎𝟎𝟑 = -0.097 < 0 ∴ 𝒙 ∈ [𝟏. 𝟎𝟎𝟎𝟑, 𝟐]

We are getting very slow convergence by Method of False position for this example.

X

f(x)

Roota

b

𝒙𝟏 𝒙𝟐Functions with Significant Curvature

𝒙𝟏𝟎= 1.1

𝑥3 =1+1.25

2= 1.125, 𝑓 𝑥3 = 𝑓 1.125 =2.147 > 0 ∴ 𝑥 ∈ [1, 1.125]

𝑥4 =1+1.125

2= 1.063, 𝑓 𝑥4 = 𝑓 1.063 =0.742 > 0 ∴ 𝑥 ∈ [1, 1.063]

𝑥5 =1+1.063

2= 1.032, 𝑓 𝑥5 = 𝑓 1.032 = 0.27 > 0 ∴ 𝑥 ∈ [1, 1.032]

𝑥6 =1+1.032

2= 1.016, 𝑓 𝑥6 = 𝑓 1.016 =0.07 > 0 ∴ 𝑥 ∈ [1, 1.016]

Part 2. Iterations for Bisection Method

𝑥1 =1+2

2= 1.5, 𝑓 𝑥1 = 𝑓 1.5 = 56.57 > 0 ∴ 𝑥 ∈ [1, 1.5]

𝑥2 =1+1.5

2= 1.25, 𝑓 𝑥2 = 𝑓 1.25 = 8.21 > 0 ∴ 𝑥 ∈ [1, 1.25]

𝑥11 =1.009+1.01

2= 1.0095, 𝑓 𝑥11 = 𝑓 1.0095 =-- 0.00008 < 0

∴ 𝑥 ∈ [1.0095, 1.01]

𝑥12 =1.0095+1.01

2= 1.0098, 𝑓 𝑥12 = 𝑓 1.0098 = - 0.002 > 0

∴ 𝑥 ∈ [1.0095, 1.0098]

∴ 𝒙 = 𝟏. 𝟎𝟎𝟗 is the approximate real root of the given equation correct up to 3 decimal

places.

𝑥8 =1.008+1.016

2= 1.012, 𝑓 𝑥8 = 𝑓 1.012 =0.027 > 0 ∴ 𝑥 ∈ [1.008, 1.012]

𝑥9 =1.008+1.012

2= 1.01, 𝑓 𝑥9 = 𝑓 1.01 = 0.005 > 0 ∴ 𝑥 ∈ [1.008, 1.01]

𝑥10 =1.008+1.01

2= 1.009, 𝑓 𝑥10 = 𝑓 1.009 = -0.006 < 0 ∴ 𝑥 ∈ [1.009, 1.01]

𝑥7 =1+1.016

2= 1.008, 𝑓 𝑥7 = 𝑓 1.008 = -0.017 < 0 ∴ 𝑥 ∈ [1.008, 1.016]

Pitfalls of the False-Position Method

Although the false-position method would seem to always be the bracketing

method of preference, there are cases where it performs poorly (as we have seen

in the last example), there are certain cases where bisection yields superior results,

particularly for functions with Significant Curvature.

The example also illustrates a major weakness of the false-position method: its

one sidedness. That is, as iterations are proceeding, one of the bracketing points

will tend to stay fixed. This can lead to poor convergence.

If f(a) and f(b) have the same sign, the function may have an even number of real

zeroes in the interval [a, b]. The False Position Method can not work for such

problems.

Open Methods

In the open methods, the method starts with one or more initialguess points. In each iteration, a new guess of the root is obtained.

Open methods are usually more efficient than bracketing methods.

They may not converge to a root.

Examples of open methods:

◦Newton - Raphson Method

◦ Fixed Point Iteration Method

Newton - Raphson Method

Newton - Raphson Method

x

f(x)

0 𝒙𝒊

𝒇(𝒙𝒊)

Root

𝒙𝒊+𝟏

𝒇(𝒙𝒊) − 𝟎

𝒙𝒊 − 𝒙𝒊+𝟏

Slope = 𝒇′(𝒙𝒊)

If the initial guess at the root is xi, a

tangent can be extended from the point

[xi, f (xi)]. The point where this tangent

crosses the x axis usually represents an

improved estimate of the root.

As in Fig., the first derivative at x is

equivalent to the slope:

Slope = 𝒇′(𝒙𝒊) =𝒇 𝒙𝒊 −𝟎

𝒙𝒊−𝒙𝒊+𝟏

⟹ 𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

Iterative formula for Newton Raphson Method

Example 8: Determine the real root of

𝒇 𝒙 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 + 𝟔𝒙 − 𝟐 = 𝟎 using Newton Raphson Method.

Solution: 𝒇 𝒙 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 + 𝟔𝒙 − 𝟐

⟹ 𝒇′ 𝒙 = 𝟏𝟓𝒙𝟐 − 𝟏𝟎𝒙 + 𝟔

𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

By Newton Raphson Method,

Let us start the iterations with the initial guess 𝒙𝟎 = 𝟎,

∴ 𝒙𝟏= 𝒙𝟎 −𝒇 𝒙𝟎𝒇′ 𝒙𝟎

= 0 -−𝟐

𝟔= 0.333

∴ 𝒙𝟐= 𝒙𝟏 −𝒇 𝒙𝟏𝒇′ 𝒙𝟏

= 0.333 -−𝟎.𝟑𝟕𝟐

𝟒.𝟑𝟑𝟑= 0.419= 𝟎. 𝟑𝟑𝟑 −

𝒇 𝟎.𝟑𝟑𝟑

𝒇′ 𝟎.𝟑𝟑𝟑

∴ 𝒙𝟑= 𝒙𝟐 −𝒇 𝒙𝟐𝒇′ 𝒙𝟐

= 0.419 -𝟎.𝟎𝟎𝟑𝟒

𝟒.𝟒𝟒𝟑= 0.4198= 𝟎. 𝟒𝟏𝟗 −

𝒇 𝟎.𝟒𝟏𝟗

𝒇′ 𝟎.𝟒𝟏𝟗

∴ 𝒙 = 𝟎. 𝟒𝟏𝟗𝟖 is one of the approximate real root of the given equation correct up to 3 decimal places.

= 0 -𝒇(𝟎)

𝒇′(𝟎)

Example 9A: Determine the lowest positive root of 𝟖𝒔𝒊𝒏(𝒙)𝒆−𝒙 − 𝟏 = 𝟎using N - R Method correct to 3 decimal places by taking 𝒙𝟎 = 𝟎.3.

Solution: 𝒇 𝒙 = 𝟖𝒔𝒊𝒏(𝒙)𝒆−𝒙 − 𝟏

⟹ 𝒇′ 𝒙 = 𝟖𝒄𝒐𝒔 𝒙 𝒆−𝒙 − 𝟖𝒔𝒊𝒏 𝒙 𝒆−𝒙 = 𝟖𝒆−𝒙(𝒄𝒐𝒔𝒙 − 𝒔𝒊𝒏𝒙)

By Newton Raphson Method,

Let us start the iterations with the initial guess 𝒙𝟎 = 𝟎. 𝟑,

∴ 𝒙𝟏= 𝒙𝟎 −𝒇 𝒙𝟎𝒇′ 𝒙𝟎

= 0.3 -𝒇(𝟎.𝟑)

𝒇′(𝟎.𝟑)= 0.108

∴ 𝒙𝟐= 𝒙𝟏 −𝒇 𝒙𝟏𝒇′ 𝒙𝟏

= 0.108 -−𝟎.𝟐𝟐𝟔

𝟔.𝟑𝟔𝟓= 0.144= 𝟎. 𝟏𝟎𝟖 −

𝒇 𝟎.𝟏𝟎𝟖

𝒇′ 𝟎.𝟏𝟎𝟖

∴ 𝒙𝟑= 𝒙𝟐 −𝒇 𝒙𝟐𝒇′ 𝒙𝟐

= 0.144 -−𝟎.𝟎𝟎𝟓𝟗

𝟓.𝟖𝟔𝟏= 0.145= 𝟎. 𝟏𝟒𝟒 −

𝒇 𝟎.𝟏𝟒𝟒

𝒇′ 𝟎.𝟏𝟒𝟒

= 0.3 -𝟎.𝟕𝟓𝟏

𝟑.𝟗𝟏𝟎

𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

Example 9B: Solve the same example by starting with 𝒙𝟎 = 𝟎.

Solution:

Let us start the iterations with the initial guess 𝒙𝟎 = 𝟎,

∴ 𝒙𝟏= 𝒙𝟎 −𝒇 𝒙𝟎𝒇′ 𝒙𝟎

= 0 -𝒇(𝟎)

𝒇′(𝟎)= 0.125

∴ 𝒙𝟐= 𝒙𝟏 −𝒇 𝒙𝟏𝒇′ 𝒙𝟏

= 0.125 -−𝟎.𝟏𝟏𝟗𝟖

𝟔.𝟏𝟐𝟓= 0.145= 𝟎. 𝟏𝟐𝟓 −

𝒇 𝟎.𝟏𝟐𝟓

𝒇′ 𝟎.𝟏𝟐𝟓

∴ 𝒙𝟑= 𝒙𝟐 −𝒇 𝒙𝟐𝒇′ 𝒙𝟐

= 0.145 -−𝟎.𝟎𝟎𝟎𝟎𝟗

𝟓.𝟖𝟒𝟖= 0.145= 𝟎. 𝟏𝟒𝟓 −

𝒇 𝟎.𝟏𝟒𝟓

𝒇′ 𝟎.𝟏𝟒𝟓

∴ 𝒙 = 𝟎. 𝟏𝟒5 is one of the approximate real root of the given equation correct up to 3 decimal places.

= 0 -−𝟏

𝟖

∴ 𝒙𝟒= 𝒙𝟑 −𝒇 𝒙𝟑𝒇′ 𝒙𝟑

= 0.145 -−𝟎.𝟎𝟎𝟎𝟎𝟗

𝟓.𝟖𝟒𝟖= 0.145= 𝟎. 𝟏𝟒𝟓 −

𝒇 𝟎.𝟏𝟒𝟓

𝒇′ 𝟎.𝟏𝟒𝟓

∴ 𝒙 = 𝟎. 𝟏𝟒5 is one of the approximate real root of the given equation correct up to 3

decimal places.

Example 10: Determine the real root of 𝟐𝒙 − 𝒍𝒐𝒈𝟏𝟎𝒙 = 7 using N - R

Method correct to 4 decimal places.

Solution: 𝒇 𝒙 = 𝟐𝒙 − 𝒍𝒐𝒈𝟏𝟎𝒙 − 7

⟹ 𝒇′(𝒙) = 𝟐 −𝟎. 𝟒𝟑𝟒𝟑

𝒙

By Newton Raphson Method,

Let us start the iterations with the initial guess 𝒙𝟎 = 𝟏,

∴ 𝒙𝟏= 𝒙𝟎 −𝒇 𝒙𝟎𝒇′ 𝒙𝟎

= 1 -𝒇(𝟏)

𝒇′(𝟏)= 4.1935

∴ 𝒙𝟐= 𝒙𝟏 −𝒇 𝒙𝟏𝒇′ 𝒙𝟏

= 4.1935 -𝟎.𝟕𝟔𝟒𝟒

𝟏.𝟖𝟗= 3.790= 4. 𝟏𝟗𝟑𝟓 −

𝒇 𝟒.𝟏𝟗𝟑𝟓

𝒇′ 𝟒.𝟏𝟗𝟑𝟓

= 1 -−𝟓

𝟏.𝟓𝟔𝟓𝟕

𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

= 𝟐𝒙 −𝒍𝒐𝒈𝒆𝒙

𝒍𝒐𝒈𝒆𝟏𝟎− 7

= 𝟐𝒙 − 𝟎. 𝟒𝟑𝟒𝟑 𝒍𝒏 𝒙 − 7

∴ 𝒙𝟑= 𝒙𝟐 −𝒇 𝒙𝟐𝒇′ 𝒙𝟐

= 3.790 -𝟎.𝟎𝟎𝟏𝟒

𝟏.𝟖𝟖𝟓𝟒= 3.7893= 𝟑. 𝟕𝟗𝟎 −

𝒇 𝟑.𝟕𝟗𝟎

𝒇′ 𝟑.𝟕𝟗𝟎

∴ 𝒙𝟒= 𝒙𝟑 −𝒇 𝒙𝟑𝒇′ 𝒙𝟑

= 3.7893 -𝟎.𝟎𝟎𝟎𝟎𝟑𝟒

𝟏.𝟖𝟖𝟓𝟒= 3.7893= 𝟑. 𝟕𝟖𝟗𝟑 −

𝒇 𝟑.𝟕𝟖𝟗𝟑

𝒇′ 𝟑.𝟕𝟖𝟗𝟑

∴ 𝒙 = 𝟑. 𝟕𝟖𝟗𝟑 is one of the approximate real root of the given equation correct up to 4

decimal places.

Example 10: Find the Newton Raphson Iteration formula for finding thevalue of cube root of any natural number N. Using it find one of the value of𝟑𝟏𝟐 correct to 3 decimal places.

Solution: Let, 𝒙 =𝟑𝑵 ⟹ 𝒙𝟑 = N

∴ 𝒇 𝒙 = 𝒙𝟑 − N = 0

∴ 𝒇′ 𝒙 = 𝟑𝒙𝟐

Where, N ∈ ℕ

By Newton Raphson Method, 𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

∴ 𝒙𝒊+𝟏= 𝒙𝒊 −𝒙𝒊

𝟑 −𝑵

𝟑𝒙𝒊𝟐 =

𝟐𝒙𝒊𝟑 +𝑵

𝟑𝒙𝒊𝟐

∴ 𝒙𝒊+𝟏=𝟐𝒙𝒊

𝟑 +𝑵

𝟑𝒙𝒊𝟐

Iterative formula for 𝟑𝑵

By N R Method

Lets find one of the value of 𝟑𝟏𝟐 by using it.

Let us start the iterations with N = 12 and the initial guess 𝒙𝟎 = 𝟏,

∴ 𝒙𝟏=𝟐𝒙𝟎

𝟑 +𝑵

𝟑𝒙𝟎𝟐

Can we take 𝒙𝟎 = 𝟎? =

𝟐+𝟏𝟐

𝟑= 4.667

∴ 𝒙𝟐=𝟐𝒙𝟏

𝟑 +𝑵

𝟑𝒙𝟏𝟐 =

𝟐 𝟒.𝟔𝟔𝟕𝟑 +𝟏𝟐

𝟑 𝟒.𝟔𝟔𝟕𝟐= 𝟐𝟏𝟓.𝟑𝟎𝟑

𝟔𝟓.𝟑𝟒𝟑= 3.295

∴ 𝒙𝟑=𝟐𝒙𝟐

𝟑 +𝑵

𝟑𝒙𝟐𝟐 =

𝟐 𝟑.𝟐𝟗𝟓𝟑 +𝟏𝟐

𝟑 𝟑.𝟐𝟗𝟓𝟐= 𝟖𝟑.𝟓𝟒𝟖

𝟑𝟐.𝟓𝟕= 2.565

∴ 𝒙𝟒=𝟐𝒙𝟑

𝟑 +𝑵

𝟑𝒙𝟑𝟐 =

𝟐 𝟐.𝟓𝟔𝟓𝟑 +𝟏𝟐

𝟑 𝟐.𝟓𝟔𝟓𝟐=

𝟒𝟓.𝟕𝟓

𝟏𝟗.𝟕𝟑𝟖= 2.318

∴ 𝒙𝟓=𝟐𝒙𝟒

𝟑 +𝑵

𝟑𝒙𝟒𝟐 =

𝟐 𝟐.𝟑𝟏𝟖𝟑 +𝟏𝟐

𝟑 𝟐.𝟑𝟏𝟖𝟐= 𝟑𝟔.𝟗𝟏

𝟏𝟔.𝟏𝟐= 2.2897

∴ 𝒙𝟔=𝟐𝒙𝟓

𝟑 +𝑵

𝟑𝒙𝟓𝟐 =

𝟐 𝟐.𝟐𝟖𝟗𝟕𝟑 +𝟏𝟐

𝟑 𝟐.𝟐𝟖𝟗𝟕𝟐= 𝟑𝟔.𝟎𝟎𝟗

𝟏𝟓.𝟕𝟑= 2.2892

∴𝟑𝟏𝟐 = 𝟐. 𝟐𝟖𝟗𝟐 correct to 3 decimal places.

Example 11: Find the Newton Raphson Iteration formula for finding the

reciprocal of any natural number N. Using it find the approximate value of𝟏

𝟏𝟗

correct to 3 decimal places.

Solution: Let, 𝒙 =𝟏

𝑵⟹

𝟏

𝒙= N

∴ 𝒇 𝒙 =𝟏

𝒙− N = 0 ∴ 𝒇′ 𝒙 = −

𝟏

𝒙𝟐

Can’t we take 𝐱 −𝟏

𝑵= 𝟎?

By Newton Raphson Method, 𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

∴ 𝒙𝒊+𝟏= 𝒙𝒊 −

𝟏𝒙𝒊

−𝑵

−𝟏𝒙𝒊

𝟐

= 𝟐𝒙𝒊 −𝑵𝒙𝒊𝟐

∴ 𝒙𝒊+𝟏= 𝟐𝒙𝒊 −𝑵𝒙𝒊𝟐 Iterative formula for

𝟏

𝑵

By N R Method

Lets find the value of 𝟏

𝟏𝟗by using it.

Let us start the iterations with N = 19 and the initial guess 𝒙𝟎 = 𝟎. 𝟎𝟏,

Just think about the interval for the root of

function 𝒇 𝒙 =𝟏

𝒙− 𝟏𝟗.

∴ 𝒙𝟏= 𝟐𝒙𝟎 − 𝟏𝟗𝒙𝟎𝟐 = 𝟐(𝟎. 𝟎𝟏) − 𝟏𝟗(𝟎. 𝟎𝟏)𝟐

Can we take 𝒙𝟎 = 𝟎?

= 𝟎. 𝟎𝟐 − 𝟎. 𝟎𝟏𝟗 = 𝟎. 𝟎𝟏𝟖𝟏

∴ 𝒙𝟐= 𝟐𝒙𝟏 − 𝟏𝟗𝒙𝟏𝟐 = 𝟐(𝟎. 𝟎𝟏𝟖𝟏) − 𝟏𝟗(𝟎. 𝟎𝟏𝟖𝟏)𝟐

= 𝟎. 𝟎𝟑𝟔𝟐 − 𝟎. 𝟎𝟎𝟔𝟐𝟑 = 𝟎. 𝟎𝟑

∴ 𝒙𝟑= 𝟐𝒙𝟐 − 𝟏𝟗𝒙𝟐𝟐= 𝟐(𝟎. 𝟎𝟑) − 𝟏𝟗(𝟎. 𝟎𝟑)𝟐

= 𝟎. 𝟎𝟔 − 𝟎. 𝟎𝟏𝟕𝟏 = 𝟎. 𝟎𝟒𝟐𝟗

∴ 𝒙𝟒= 𝟐𝒙𝟑 − 𝟏𝟗𝒙𝟑𝟐 = 𝟐(𝟎. 𝟎𝟒𝟐𝟗) − 𝟏𝟗(𝟎. 𝟎𝟒𝟐𝟗)𝟐

= 𝟎. 𝟎𝟖𝟓𝟖 − 𝟎. 𝟎𝟑𝟓 = 𝟎. 𝟎𝟓𝟎𝟖

∴ 𝒙𝟓= 𝟐𝒙𝟒 − 𝟏𝟗𝒙𝟒𝟐 = 𝟐(𝟎. 𝟎𝟓𝟎𝟖) − 𝟏𝟗(𝟎. 𝟎𝟓𝟎𝟖)𝟐

= 𝟎. 𝟏𝟎𝟏𝟔 − 𝟎. 𝟎𝟒𝟗 = 𝟎. 𝟎𝟓𝟐𝟔

∴ 𝒙𝟔= 𝟐𝒙𝟓 − 𝟏𝟗𝒙𝟓𝟐 = 𝟐(𝟎. 𝟎𝟓𝟐𝟔) − 𝟏𝟗(𝟎. 𝟎𝟓𝟐𝟔)𝟐

= 𝟎. 𝟏𝟎𝟓𝟐 − 𝟎. 𝟎𝟓𝟐𝟔= 𝟎. 𝟎𝟓𝟐𝟔

∴𝟏

𝟏𝟗= 𝟎. 𝟎𝟓𝟐𝟔 correct to 3 decimal places.

Pitfalls of N R Method f(x) is continuous and the first derivative is known.

An initial guess x0 such that f ’(x0) ≠ 0.

If the initial guess of the root is far from the root the method may notconverge i.e., Convergence Depends on a Good Initial Guess.

X

f(x)

0x

1x2x 0x

1x

N R Method can be divergent or can be produce oscillatory results.

Sometimes, it may converge very slowly.

1:of roots positive theFind :Example 10 xf(x)

5.0 use 10

19

10

1

o

i

iii x

x

xxx:FormulaR -N

Iteration

0

1

2

3

4

5

.

.

38

∞

x

0.5

51.65

46.485

41.8365

37.65285

33.887565

1.083

1.0000000

Fixed Point Iteration Method

Fixed Point Iteration

X

Root

x is a root if f(x) = 0,

xexf x

xe x 0

Fixed Point Iteration

X

X

f1(X)

f2(X)f1(X) f2(X)

root

root

xe x 0

xex

Fixed Point Iteration

x is a root if,

X

f1(X)

f2(X)

root

f2(X)

xex

f1(X)

f1(x) = f2(x)

Convergence Examples

Convergent staircase pattern Convergent spiral pattern

Divergence Example

Divergent staircase pattern Divergent spiral pattern

Example 14: Using the method of fixed point iteration, find the roots of 𝒙𝟑−𝒙− 𝟏 = 𝟎 correct to four decimal places.

Solution: To check the convergence criteria lets find the interval first,𝑥 = 1 ⟹ 𝑓 1 = −1 < 0

𝑥 = 2 ⟹ 𝑓 2 = 5 > 0

∴ 𝑥 ∈ [1, 2]

Lets think to take iterative formula, 𝑥 = 𝑥3 − 1∴ 𝑔 𝑥 = 𝑥3 − 1 ⟹ 𝑔′ 𝑥 = 3𝑥2

As, 𝑔′(𝑥) > 1, 𝑓𝑜𝑟 ∀𝑥 ∈ [1,2]∴ 𝑥 = 𝑥3 − 1 will not give the convergent iteration.

Lets think to take next iterative formula, 𝑥 = (𝑥 + 1)1/3

∴ 𝑔 𝑥 = (𝑥 + 1)1/3⟹ 𝑔′ 𝑥 =1

3(𝑥 + 1)2/3

As, 𝑔′(𝑥) < 1, 𝑓𝑜𝑟 ∀𝑥 ∈ [1,2]

∴ 𝑥 = (𝑥 + 1)1/3 will give the convergent iteration.

Hence, fixed point iterative formula be,

𝑥𝑖+1 = (𝑥𝑖 + 1)1/3, 𝑖 = 0,1,2, …

Let, initial approximation be 𝑥0 = 1.5,

𝑥1 = (𝑥0 + 1)1/3= (1.5 + 1)1/3= 1.3572

𝑥2 = (𝑥1 + 1)1/3= (1.3572 + 1)1/3= 1.3309

𝑥3 = (𝑥2 + 1)1/3= (1.3309 + 1)1/3= 1.3259

𝑥4 = (𝑥3 + 1)1/3= (1.3259 + 1)1/3= 1.3249

𝑥5 = (𝑥4 + 1)1/3= (1.3249 + 1)1/3= 1.3248

𝑥6 = (𝑥5 + 1)1/3= (1.3248 + 1)1/3= 1.3247

𝑥7 = (𝑥6 + 1)1/3= (1.3247 + 1)1/3= 1.3247

∴ 𝒙 =1.3247 is the approximate real root of the given equation correct up to 4

decimal places.

Example 15: Using the method of fixed point iteration, find the roots of 𝟑𝒙 + 𝟐 𝒔𝒊𝒏𝒙 = 𝒆𝒙 correct to three decimal places.

Solution: To check the convergence criteria lets find the interval first,𝑥 = 0 ⟹ 𝑓 0 = −1 < 0

𝑥 = 1 ⟹ 𝑓 1 = 1.965 > 0

∴ 𝑥 ∈ [0, 1]

Lets iterative formula be, 𝑥 = ln(3𝑥 + 2𝑠𝑖𝑛𝑥)

∴ 𝑔 𝑥 = ln(3𝑥 + 2𝑠𝑖𝑛𝑥) ⟹ 𝑔′ 𝑥 =3 + 2𝑐𝑜𝑠𝑥

3𝑥 + 2𝑠𝑖𝑛𝑥As, g′ 𝑥 is undefined for 𝑥 = 0∴ 𝑥 = ln(3𝑥 + 2𝑠𝑖𝑛𝑥) will not give the convergent

iteration.

Lets think to take next iterative formula, 𝑥 =1

3(𝑒𝑥 − 2𝑠𝑖𝑛𝑥)

∴ 𝑔 𝑥 =1

3(𝑒𝑥 − 2𝑠𝑖𝑛𝑥) ⟹ 𝑔′ 𝑥 =

1

3(𝑒𝑥 − 2𝑐𝑜𝑠𝑥)

As, 𝑔′(𝑥) < 1, 𝑓𝑜𝑟 ∀𝑥 ∈ [0,1]

∴ 𝑥 =1

3(𝑒𝑥 − 2𝑠𝑖𝑛𝑥) will give the convergent iteration.

Hence, fixed point iterative formula be,

𝑥𝑖+1 =1

3𝑒𝑥𝑖 − 2𝑠𝑖𝑛𝑥𝑖 , 𝑖 = 0,1,2, …

Let, initial approximation be 𝑥0 = 0.5,

𝑥1 =1

3𝑒𝑥0 − 2𝑠𝑖𝑛𝑥0 =

1

3𝑒0.5 − 2sin(0.5) = 0.23

𝑥2 =1

3𝑒𝑥1 − 2𝑠𝑖𝑛𝑥1 =

1

3𝑒0.23 − 2sin(0.23) = 0.2676

𝑥3 =1

3𝑒𝑥2 − 2𝑠𝑖𝑛𝑥2 =

1

3𝑒0.2676 − 2sin(0.2676) = 0.2593

𝑥4 =1

3𝑒𝑥3 − 2𝑠𝑖𝑛𝑥3 =

1

3𝑒0.2593 − 2sin(0.2593) = 0.2611

𝑥5 =1

3𝑒𝑥4 − 2𝑠𝑖𝑛𝑥4 =

1

3𝑒0.2611 − 2sin(0.2611) = 0.2607 ≈ 0.261

∴ 𝒙 = 𝟎. 𝟐𝟔𝟏 is the approximate real root of the given equation correct up to 3 decimal

places.

Convergence Criterion of N-R Method

𝒙𝒊+𝟏 = 𝒈 𝒙𝒊

𝒙𝒊+𝟏 = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

⟹ 𝒈(𝒙𝒊) = 𝒙𝒊 −𝒇 𝒙𝒊𝒇′ 𝒙𝒊

⟹ 𝒈(𝒙) = 𝒙 −𝒇 𝒙

𝒇′ 𝒙⟹ 𝒈′(𝒙) = 𝟏 −

𝒇′(𝒙)𝒇′ 𝒙 − 𝒇 𝒙 𝒇′′(𝒙)

𝒇′(𝒙)𝟐

⟹ 𝒈′(𝒙) =𝒇 𝒙 𝒇′′(𝒙)

𝒇′(𝒙)𝟐

∴ 𝒈′(𝒙) < 𝟏 ⟹ 𝒇 𝒙 𝒇′′ 𝒙 < 𝒇′ 𝒙 𝟐Convergence Criterion

for N - R Method

Next Chapter:Numerical Solution of System of Linear Equations