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Polynomial An expression of the form f ( ) x = a x 0 n + a x 1 n −^1 + a x 2 n −^2 + ...+ an (^) − 1 x + an where n is a positive integer and a 0 (^) , a 1 (^) , a 2 + .... an are real constants, such type of expression is called an nth degree polynomial in x if a 0 (^) ≠ 0
Algebraic equation:
An equation f(x)=0 is said to be the algebraic equation in x if it is purely a polynomial in x. For example x^5^ + x^4 + 3 x^2 + x − 6 = 0 It is a fifth order polynomial and so this equation is an algebraic equation. 3 6 4 3 2 4 2
x x x y y y y polynomial in y t t polynomail in t
These all are the examples of the polynomial or algebraic equations.
Some facts
- Every equation of the form f(x)=0 has at least one root ,it may be real or complex.
- Every polynomial of nth degree has n and only n roots.
- If f(x) =0 is an equation of odd degree, then it has at least one real root whose sign is opposite to that of last term. 4.If f(x)=0 is an equation of even degree whose last term is negative then it has at least one positive and at least one negative root.
Transcendental equation
An equation is said to be transcendental equation if it has logarithmic, trigonometric and exponential function or combination of all these three. For example e x − 5 x − 3 = 0 it is a transcendental equation as it has an exponential function
2
sin 0 ln sin 0 2sec tan 0
x
x
e x x x x x e
These all are the examples of transcendental equation.
Root of an equation
For an equation f(x) =0 to find the solution we find such value which satisfy the equation f(x)=0,these values are known as the roots of the equation.
A value a is known as the root of an equation f(x) =0 if and only if f (a) =0. docsity.com
Properties of an Algebraic equation
- Complex roots occur in the pairs. That is ,If (a+ib ) is a root of f(x)=0 then (a-ib ) is also a root of the equation
- if x=a is a root of the equation f(x)=0 a polynomial of nth degree ,then (x-a) is a factor of f(x) and by dividing f(x) by (x-a) we get a polynomial of degree n-1.
Descartes rule of signs This rule shows the relation ship between the signs of coefficients of an equation and its roots. “The number of positive roots of an algebraic equation f(x) =0 with real coefficients can not exceed the number of changes in the signs of the coefficients in the polynomial f(x) =0.similarly the number of negative roots of the equation can not exceed the number of changes in the sign of coefficients of f (-x) =0”
Consider the equation x^3^ − 3 x^2 + 4 x − 5 = 0 here it is an equation of degree three and there are three changes in the signs First +ve to –ve second –ve to +ve and third +ve to –ve so the tree roots will be positive Now f ( − x ) = − x^3^ − 3 x^2 − 4 x − 5 so there is no change of sign so there will be no negative root of this equation.
Intermediate value property
If f(x) is a real valued continuous function in the closed interval a^ ≤^ x^ ≤^ b if f(a) and f(b)
have opposite signs once; that is f(x)=0 has at least one root β such that a ≤ β≤ b
Simply If f(x)=0 is a polynomial equation and if f(a) and f(b) are of different signs ,then f(x)= must have at least one real root between a and b.
Numerical methods for solving either algebraic or transcendental equation are classified into two groups
Direct methods
Those methods which do not require any information about the initial approximation of root to start the solution are known as direct methods. The examples of direct methods are Graefee root squaring method, Gauss elimination method and Gauss Jordan method. All these methods do not require any type of initial approximation.
Iterative methods
These methods require an initial approximation to start.
between both these points ,this method can be used for both transcendental and algebraic equations. Consider the equation
(0) 1
(1) 3 1 sin(1 180 ) 3 1 0.84147 1.
f
f
= − + × = − + =
Here f(0) and f(1) are of opposite signs making use of intermediate value property we infer that one root lies between 0 and 1. So in analytical method we must always start with an initial interval (a,b) so that f(a) and f(b) have opposite signs.
Bisection method (Bolzano)
Suppose you have to locate the root of the equation f(x)=0 in an interval say ( x 0 (^) , x 1 ),let f ( x 0 )and f ( x 1 )are of opposite signs such that f ( x 0 (^) ) f ( x 1 ) < 0
Then the graph of the function crossed the x-axis between x 0 (^) and x 1 which exists the existence of at least one root in the interval ( x 0 (^) , x 1 ).
The desired root is approximately defined by the mid point 2 0 1 2 x =^ x^ + x if f ( x 2 ) = 0 then x 2 is the root of the equation otherwise the root lies either between x 0 (^) and x 2 or x 1 (^) and x 2
Now we define the next approximation by 3 0 2 2 x =^ x^ + x provided f ( x 0 (^) ) f ( x 2 ) < 0 then
root may be found between x 0 (^) and x 2
If provided f ( x 1 (^) ) f ( x 2 ) < 0 then root may be found between x 1 (^) and x 2 by 3 1 2 2 x =^ x^ + x
Thus at each step we find the desired root to the required accuracy or narrow the range to half the previous interval. This process of halving the intervals is continued in order to get smaller and smaller interval within which the desired root lies. Continuation of this process eventually gives us the required root.
NOTE: In numerical analysis all the calculation are carried out in radians mode and the value of pi is supposed to be 3.
Example Solve x^3^ − 9 x + 1 = 0 for the root between x=2 and x=4 by bisection method
Solution: Here we are given the interval (2,4) so we need not to carry out intermediate value property to locate initial approximation. Here
f ( ) x = 3 x − 1 + sin x = 0
3 3 3
f x x x now f f here f f so root lies between and
0 1 2 3
3 3
x x x f here f f so the root lies between nad x f so the root lies between and as f f
4
5 6
x now similarly x and x and the process is continued untill the desired accuracy is obtained
n xn f ( xn )
2 3 1.
3 2.5 -5.
4 2.75 -2.
5 2.875 -1.
6 2.9375 -0.
When to stop the process of iteration? Here in the solved example neither accuracy is mentioned and nor number of iteration is mentioned by when you are provided with the number of iteration then you will carry those number of iteration and will stop but second case is accuracy like you are asked to
2 2
2
( ) cos 2 3 1 (1.2) 1.2 cos1.2 2(1.2) 3(1.2) 1 1.2(0.3623) 2(1.44) 3.6 1 (1.3) 1.3cos1.3 2(1.3) 3(1.3) 1 1.3(0.2674) 2(1.69) 9.3 1 (1.2) (1.3) 0
f x x x x x f
f
as f f
2 2
3
(1.25) 1.25cos1.25 2(1.25) 3(1.25) 1 1.25(0.3153) 2(1.5625) 3.75 1 (1.25) (1.3) 0 1.25 1.3 (^1) 2
so the root lies between both
x
f
as f f so the root lies between both
x
2
4
(1.275) 1.275cos1.275 2(1.275) 3(1.275) 1 1.275(0.2915) 2(1.6256) 3.825 1 (1.25) (1.275) 0 1.25 1.275 (^) 1. 2 (1.2625) 1.2625cos1.
f
as f f so the root lies between both
x
f
5 2
(1.25625) 1.25625cos1.25625 2(1.25625) 3(1.
as f f so the root lies between both
x
f
6
as f f so the root lies between both
x
