REGULA - FALSI METHOD | ENGINEERING MATHEMATICS | FINDIND ROOTS OF AN EQUATION

INTRODUCTION:

Regula-Falsi method is also used for finding approximate roots of an equation .It is very similar to bisection method.

                If you don't read about bisection method OR forgot the bisection method just click here to read now.

This is also called False Position Method.

THEORY:

As before (in Bisection Method), for a given continuous function f(x) we assume that f (a) and f (b) have opposite signs ( a = lower guess, b = upper guess). This condition satisfies the Balzano's Theorem for continuous function.

Note::Theorem (Bolzano) : If the function f(x) is continuous in [a, b] and f(a)f(b) < 0 (i.e. f(x) has opposite signs signs at a and b) then a value c ∈ (a, b) exists such that f(c) = 0.

Now after this bisection method used the midpoint of the interval [a, b] as the next iterate to converge towards the root of f(x).

A better approximation is obtained if we find the point (c, 0) where the secant line L joining the points (a, f (a)) and (b, f (b)) crosses the x-axis (see the image below). To find the value c, we write down two versions of the

slope m of the line L:

We first use points (a, f (a)) and (b, f (b)) to get equation 1 (below), and then use the points (c, 0) and (b, f (b)) to get equation 2 (below). Equating these two equation we get equation 3 (below) which is easily solved for c to get equation 4 (below):

The three possibilities are the same as before in the bisection method:

• If f (a) and f (c) have opposite signs, a zero lies in [a, c].
• If f (c) and f (b) have opposite signs, a zero lies in [c, b].
• If f (c) = 0, then the zero is c.

A short GIF for how this method works:

Working Rules:

1. Find Points x0 and x1such that x0< x1and f(x0).f( x1) < 0
2. Take the interval [x0, x1] and find next value x2= x0-f( x0).(( x1- x0)/f( x1)-f( x0))
3. If f(x2)=0 then x2is an exact root.
4. If f(x1).f(x2) < 0 then x1= x2
5. If f(x0).f(x2) <0 then x0= x2
6. Repeat the step 2-5 until f(xi)=0 or f(x0)≤ Accuracy

Solve: Find a root of an equation f(x)=x3-x-1 using False Position method

Solution:
Here x3-x-1=0

Let f(x)=x3-x-1

x	0	1	2
f(x)	-1	-1	5

1st iteration :

Here f(1)=-1<0 and f(2)=5>0

∴ Now, Root lies between x0=1 and x1=2

x2=1-(-1)⋅(2-1)/(5-(-1))

x2=1.16667

f(x2)=f(1.16667)=-0.5787<0

2nd iteration :[ FROM HERE THIS LEAVES FOR READERS AND HOPE ALL ITERATION WILL BE SOLVED ]

LASTLY WE GOT

[HERE n= No. of Iteration , x0 = value of 1st Guess point and f( x0 )= value of Fuction at x0 ]
Approximate root of the equation x3-x-1=0 using False Position method is 1.32464

Newton-Raphson Method ---Solution of Polynomial And Transcendental equation| Engineering Mathematics

Introduction:
≫In the previous session we read about Bisection method which is used to solve the equation for finding root.
if you are not read that blog i suggest you to read that blog first
Click Here to read.

Newton-Raphson Method:

Concept:

We will choose a +ve point much nearer to cut of the curve to X-axix . Let that point is x_n and it intersect the curve at (x_n,∱(x_n)) .

              If we draw a tagent at the point (x_n,∱(x_n)) then the slope will be ∱'(x_n) and cut the X-axis at x_n+1.

     Similarly,the point x_n+1 cuts the curve at (x_n+1,∱(x_n+1)) and if we draw a tangent at there then the slope will be ∱'(x_n+1) and the tangent cut the X-axix at x_n+2 .

         And The point x_n+2 cuts the curve at (x_n+2,∱(x_n+2)). 

We can see that at every step we move towards the root point of ∱(x)=0.

This method is very easy and less-effort than Bisection-Method.

Working Rule:

1. Given ∱(x)=0 ------(1)
   Find initial root x₀ such that ∱(x)~0
[ i.e; x₀ is near to the root ∱(x)=0]

2. Find ∱(x₀) and ∱'(x₀)

3. First approximate root by Newton-Raphson method:
            $x_{1=x_{0\mathrm{}}}- \frac{∱\left(x_{0\mathrm{}}\right)}{∱\text{'}\left(x_{0\mathrm{}}\right)}$      

Find ∱(x₁) and ∱'(x₁)

4. Similarly, Second approximate root is

$x_{2=x_{1\mathrm{}}}- \frac{∱\left(x_{1\mathrm{}}\right)}{∱\text{'}\left(x_1\right)}$              

5.Also Third approximate root is

          $x_{3=x_{2\mathrm{}}}- \frac{∱\left(x_{2\mathrm{}}\right)}{∱\text{'}\left(x_{2\mathrm{}}\right)}$                 

General Formula:


$x_{n =x_{n-1 }}- \frac{∱\left(x_{n-1}\right)}{∱\text{'}\left(x_{n-1}\right)}$

Example

QnS. ∱(x) = x - 2 + lnx has a root near x =1.5 .Use Newton-Raphson Method to obtain a better estimate 2 digit after decimal point.

Ans.

In this question we don't need to figure out the x₀, because it has given that the root is lies near x=1.5 .

Here x₀ =1.5 and ∱(1.5) = -0.5 + ln(1.5) = -0.0945

∱'(x)= 1+(1/x)

⇒ ∱'(1.5) = 5/3

Hence using the general formula

x₁=1.5567

The Newton-Raphson formula can be used again.

so ∱(x₁) = ∱(1.5567) = -0.0007

∱'(x₁) = ∱'(1.5567) = 1.6424

{ Calculation Part is skipped for Readers}

x₂= 1.5571

⇒We will note that the 2digits after decimal point of x₁ is same as x₂ .So

our answer will be x₂.

Solution of Polynomial And Transcendental equation - Bisection Method | Engineering Mathematics

Introduction:

Before Expressing the Bisection Method.We should discuss about the equations and graphs.

            (Graph of an equation)

This is a graph of an equation. As we all know equations are of two types

1.Polynomial Equation

2.Transcendental Equation 

A polynomial equation of the form

is called an Algebraic equation. For Example

An equation which contains polynomials,trigonometry functions,logarithmic functions,exponential functions etc.,is called a Transcendental equation.For example

are Transcendental equations.
>>A polynomial equation of degree n will have exactly n roots,real or complex , simple or multiple. A transcendental equation may have one root  or no root or infinite numbers of roots depending on the form of  ∱(x) .

The Methods of finding the roots of ∱(x) =0 are classified as

1.Direct Method:

Direct methods give the exact values of all the roots in a finite number of steps.
                               In these methods we start with one or two initial approximation of the root and obtain a sequence of approximation X0 ,X1 ,...... ,Xk  which in the limit as K ￫  ∞  converge to the exact root X =a.
                                                       There are no direct methods for solving higher degree algebraic equation or transcendental equation.

2.Numerical Method:

Numerical methods are based are based on the idea of successive approximations.
                           In these method , we first find an interval in which the root lies.If 'a' and 'b' are two number such that ∱(a) and ∱(b) have opposite sign,then a root ∱(x)=0 lies in between 'a' and 'b' . We will discuss few important Numerical Methods to find root of ∱(x)=0 .

2.1 Bisection Method:
Bisection method is a time taking method which needs a much effort.
Working Rule:
(i) Let ∱(x)=0 be the given equation . Find 'a' and 'b' such that ∱(a).∱(b)<0
we consider ∱(a)<0 and ∱(b)>0.
{One of them must take +ve and other should -ve}

(ii)Find first approx. root using bisection method as
                                    X1= (a+b)/2
Calculate ∱(x1) and examine the sign.

(iii)If ∱(x1) <0 ➡️ root lies between X1 and 'b'.
Similarly 2nd approx root is given by
                                  X2=(X1 +b)/2
(iv)If ∱(x)>0 ➡️ root lies between 'a' and X1 then 2nd approx. root is given by
                                  X2=(X1 +a)/2
Calculate ∱(X2) and repeat step (iii) & (iv) until we get the required accuracy of the root.

QNs. Find The real roots of the equation x³ -x -4=0.using bisection method correct upto 3 decimal places?
ANS:
Here  ∱(x)=x³ -x -4=0

Find 'a' and 'b':
∱(0)=-4 < 0
∱(1) =-4 <0
∱(2)=2>0
{we go much closer to root thats why we can find the root with less effort}
∱(1.5)=-2.125<0
∱(1,6)=-1.504<0
∱(1.7)=-0.787<0
∱(1.8)=0.032>0
{We can notice that ∱(1.7) and ∱(1,8) are much closure to zero }

Let a=1.7 and b=1.8


First Approximate root using bisection method:

X1= (a+b)/2=1.75
➡️∱(1.75)=-0.34<0
Hence root lies between 1.75 and 1.8

Second Approximate root:
X2=(1.75 +1.8)/2=1.775
➡️∱(1.775)= -0.182<0
Hence root lies between 1.775 and 1.8.

Third Approximate root:
X3=(1.775+1.8)/2 = 1.7875
➡️∱(1.7875)= -0.076<0
Hence the root lies between 1.7875 and 1.8

Fourth Approximate root:
X4=(1.7875 +1.8)/2 = 1.79375
∱(1.79375)= -0.022<0
Hence the root lies between 1.79375 and 1.5

Fifth Approximate root:
X5=(1.79375 +1.8)/2=1.796875
∱(1.796875)=0.0048>0
Hence root lies between 1.79375 and 1.796875

Sixth Approximate root:
X6=(1.79375+1.796875)/2=1.795312
∱(1.795312)=-0.0087<0
Hence the root lies between 1.795312 and 1.796875

Seventh Approximate root:
X7=(1.795312+1.796875)/2=1.796093

If  new root's 3digit after decimal is matched with any of the Previous two root's 3digit after decimal then we concluded that we found the answer.

So X7=1.796093 matched with X5=1.796875.Now X7 is our approximate root.