Unit-6
Numerical solution of Transcendental Equations, System of Linear equations and Expansion of Functions.
There are two types of equations Linear and Non linear equations. Linear equations are those in which dependent variable y is directly proportional to independent variable x and is of degree one. On the other hand non linear equation are those in which y does not directly proportional to x and of degree more than one.
Ex: +b, where a and b are constant is a linear equation.
is a non linear equation.
Algebraic Equation: If f(x) is a pure polynomial, then the equation is called an algebraic equation in x.
Ex:
Transcendental Equation: If f(x) is an expression contain function as trigonometric, exponential and logarithmic etc. Then is called transcendental equation.
Ex
Non –linear equation can be solved by using various analytical methods. The transcendental equations and higher order algebraic equations are difficult to solve even sometime are impossible. Finding solution of equation means just to calculate its roots.
Numerical methods are often repetitive in nature. They consist of repetitive calculation of the same process; where in each step the result of preceding values are used (substitute). This is known as iteration process and is repeated till the result is obtained to desired accuracy.
The analytical methods used to solve equation; exact value of the root is obtained whereas in numerical method approximate value is obtained.
The numerical methods to find roots of non linear equations are:
- Regula-Falsi Method (Method of False position)
- Newton Raphson Method
Some important theorem:
Theorem1: If f(x) is continuous in ,and if f(a) anf f(b) are of opposite signs, then for atleast one numberµ such that
Theorem2:(Rolle’s theorem) If f(x) is continuous in , f’(x) exists in and f(a) = f(b)=0 , then there exist one numberµ such that such that
Theorem3: (Intermediate value theorem) let f(x) be a continuous function in [a,b] and let k be any number between f(a) and f(b). Then there exists a µ in(a,b) such that f(µ) = k.
Theorem 4: (Mean Value theorem) Let f(x) be continuous in [a,b] and f’(x) exist in (a,b), then there exists at least one value of x, say µ, between a and b such that
6.1.1. Regula Falsi Method:
This is the oldest method of finding the approximate numerical value of a real root of an equation.
In this method we suppose that and are two points where and are of opposite sign .Let
Hence the root of the equation lies between and and so,
The Regula Falsi formula
Find is positive or negative. If then root lies between and or if then root lies between and similarly we calculate
Proceed in this manner until the desired accurate root is found.
Example1 Find a real root of the equation near, correct to three decimal place by the Regula Falsi method.
Let
Now,
And also
Hence the root of the equation lies between and and so,
By Regula Falsi Method
Now,
So the root of the equation lies between 1 and 0.5 and so
By Regula Fasli Method
Now,
So the root of the equation lies between 1 and 0.63637 and so
By Regula Fasli Method
Now,
So the root of the equation lies between 1 and 0.67112 and so
By Regula Fasli Method
Now,
So the root of the equation lies between 1 and 0.63636 and so
By Regula Fasli Method
Now,
So the root of the equation lies between 1 and 0.68168 and so
By Regula Fasli Method
Now,
Hence the approximate root of the given equation near to 1 is 0.68217
Example2 Find the real root of the equation
By the method of false position correct to four decimal places
Let
By hit and trail method
0.23136 > 0
So, the root of the equation lies between 2 and 3 and also
By Regula Falsi Mehtod
Now,
So, root of the equation lies between 2.72101 and 3 and also
By Regula Falsi Mehtod
Now,
So, root of the equation lies between 2.74020 and 3 and also
By Regula Falsi Mehtod
Now,
So, root of the equation lies between 2.74063 and 3 and also
By Regula Falsi Mehtod
Hence the root of the given equation correct to four decimal places is 2.7406
Example3 Apply Regula Falsi Method to solve the equation
Let
By hit and trail
And
So the root of the equation lies between and also
By Regula Falsi Mehtod
Now,
So, root of the equation lies between 0.60709 and 0.61 and also
By Regula Falsi Method
Now,
So, root of the equation lies between 0.60710 and 0.61 and also
By Regula Falsi Method
Hence the root of the given equation correct to five decimal place is 0.60710.
6.1.2. Newton-Raphson Method:
Let be the approximate root of the equation.
By Newton Raphson formula
In general,
Where n=1, 2, 3…… we keep on calculating until we get desired root to the correct decimal places.
Example1 Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places:.
Given
By Newton Raphson Method
=
=
The initial approximation is in radian.
For n =0, the first approximation
For n =1, the second approximation
For n =2, the third approximation
For n =3, the fourth approximation
Hence the root of the given equation correct to five decimal place 2.79838.
Example2 Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5
Let
The initial approximation
By Newton Raphson Method
For n =0, the first approximation
For n =1, the second approximation
For n =2, the third approximation
For n =3, the fourth approximation
Hence the root of the equation correct to three decimal places is 4.5579
Example3 Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:
Let
By Newton Raphson Method
Let the initial approximation be
For n=0, the first approximation
For n=1, the second approximation
For n=2, the third approximation
Since therefore the root of the given equation correct to four decimal places is -2.9537
The standard a system of n linear equation in n unknown is
(1)
… …… …. …
6.2.1.Gauss Jacobi’s Iteration method :
Let us consider the system of simultaneous linear equation
(1)
The coefficients of the diagonal elements are larger than the all other coefficients and are non zero. Rewrite the above equation we get
(2)
Take the initial approximation we get the values of the first approximation of .
By the successive iteration we will get the desired the result.
Example1Use Jacobi’s method to solve the system of equations:
Since
So, we express the unknown with large coefficient in terms of other coefficients.
(1)
Let the initial approximation be
2.35606
0.91666
1.932936
0.831912
3.016873
1.969654
3.010217
1.986010
1.988631
0.915055
1.986532
0.911609
1.985792
0.911547
1.98576
0.911698
Since the approximation in ninth and tenth iteration is same up to three decimal places, hence the solution of the given equations is
Example2 Solve by Jacobi’s Method, the equations
Given equation can be rewrite in the form
… (i)
..(ii)
..(iii)
Let the initial approximation be
Putting these values on the right of the equation (i), (ii) and (iii) and so we get
Putting these values on the right of the equation (i), (ii) and (iii) and so we get
Putting these values on the right of the equation (i), (ii) and (iii) and so we get
0.90025
Putting these values on the right of the equation (i), (ii) and (iii) and so we get
Putting these values on the right of the equation (i), (ii) and (iii) and so we get
Hence solution approximately is
Example3Use Jacobi’s method to solve the system of the equations
Rewrite the given equations
(1)
Let the initial approximation be
1.2
1.3
0.9
1.03
0.9946
0.9934
1.0015
Hence the solution of the above equation correct to two decimal places is
6.3.1
6.2.3.Gauss Seidel method:
This is the modification of the Jacobi’s Iteration.
Let us consider the system of simultaneous linear equation
(1)
The coefficients of the diagonal elements are larger than the all other coefficients and are non zero. Rewrite the above equation we get
(2)
Take the initial approximation we get the values of the first approximation of .
As above in Jacobi’s Iteration, we take first approximation as
and put in the right hand side of the first equation of (2) and let the result be .
Now we put right hand side of second equation of (2) and suppose the result is
Now put in the RHS of third equation of (2) and suppose the result be
The above method is repeated till the values of all the unknown are found up to desired accuracy.
Example1 Use Gauss –Seidel Iteration method to solve the system of equations
Since
So, we express the unknown of larger coefficient in terms of the unknowns with smaller coefficients.
Rewrite the above system of equations
(1)
Let the initial approximation be
3.14814
2.43217
2.42571
2.4260
Hence the solution correct to three decimal places is
Example2 Solve the following system of equations
By Gauss-Seidel method.
Rewrite the given system of equations as
(1)
Le t the initial approximation be
Thus the required solution is
Example3 Solve the following equations by Gauss-Seidel Method
Rewrite the above system of equations
(1)
Let the initial approximation be
Hence the required solution is
Expansion of some well known series:
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
6.3.1Taylor’s theorem:
If (i) f(x) and its first (n-1) derivative be continuous in [a, a+h],
(ii) exist for every value of x in (a, a+h), then there is at least one number such that
This is called Taylor’s theorem with Lagrange’s form of remainder
Taylor’s Series:
If can be expanded as an infinite series, then
If possesses derivative of all orders and the remainder .
Corollary: Taking and in equation (i) we get
Taking in above we get Maclaurin’s series.
Example1: Expand the polynomial in power of , by Taylor’s theorem.
Let .
Also
Then
Differentiating with respect to x.
Again differentiating with respect to x the above function.
Again differentiating with respect to x the above function.
Also the value of above functions at x=2 will be
By Taylor’s theorem
On substituting above values we get
Example2: Expand in power of
Let
Also
Differentiating f(x) with respect to x.
Again differentiating f(x) with respect to x.
Again differentiating f(x) with respect to x.
Also the value of above functions at x=1 will be
By Taylor’s theorem
On substituting above values we get
=
Example3: Expand in power of. Hence find the value of correct to four decimal places.
Let
And .
Differentiating with respect to x.
Again differentiating with respect to x.
Again differentiating with respect to x.
Again differentiating with respect to x.
Also the value of above functions at will be
By Taylor’s theorem
On substituting above values we get
At
.
6.3.2.Maclaurin’s theorem:
This is a particular case of Taylor’s theorem in which a=0 and h=x in Taylor’s theorem.
If f(x) can be expanded as an infinite series, then
Where the remainder is
Example1:If using Taylor’s theorem, show that for .
Deduce that
Let then
Differentiating with respect to x.
.Then
Again differentiating with respect to x.
Then
Again differentiating with respect to x.
Then
By Maclaurin’s theorem
Substituting the above values we get
Since
Hence
Example2: Prove that
Let
Differentiating with respect to x.
Again differentiating with respect to x.
Again differentiating with respect to x.
Again differentiating with respect to x.
and so on.
Putting , in above derivatives we get
so on.
By Maclaurin’s theorem
+………
Substituting the above values we get
Example3: Prove that
Let
Differentiating above function with respect to x.
Again differentiating above function with respect to x.
Again differentiating above function with respect to x.
Again differentiating above function with respect to x.
Putting , in above derivatives we get
so on.
By Maclaurin’s theorem
+………
Substituting the above values we get
Example: Expansion of Some standard series
Let and
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
2.
Let and
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
3.
Let and
Differentiating above function with respect to x.
.
By Maclaurin’s theorem
+………
Substituting the above values we get
4.
Let and also
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
5.
Let and also
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
6.
Let and also
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
7.
Let
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
++………
+………
8.
Let
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
+………
+………
9.
Let
Differentiating above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get
+………
+………)
10.
Let
Differentiating the above function with respect to x.
By Maclaurin’s theorem
+………
Substituting the above values we get