Polynomial variation
Introduction :
A polynomial that contains only one variable,we take it as x, is known as a polynomial in the variable x.
Thus, 5x2 + 13x – 9 is a polynomial in the variable x.
y3 + 7y – 19 is a polynomial in the variable y.
It is customary to write the terms of a polynomial in decreasing order of the exponents of the variables. This is called the standard form of a polynomial. The term that does not contain the variable is written at the end. It is known as the constant term.
Usage of the polynomial variation:
Solve the f(x) = `(1)/(x-3)` , x = 4. Find the degree of polynomial up to three.
Solution:
f(x) = (1) /( x - 3) f (4) = 2
f ‘(x) = (-1) /(( x - 3)^2) f ‘(4) = -1
f ‘’(x) = (2) /(( x - 4)^3) f ‘’(4) = 2
To find the Taylor polynomial:
General form of the Taylor polynomials is,
Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n
Here, we calculate up to third degree of Taylor polynomial.
Pn(x) = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!) + f’’’ (4) ((x – a)3/ 3!)
= 2 + - (1) (x - 4) + 2((x – 4)2 / 2!) + (-6) (x – 4)3/ 6
= 2 - x + 4 + (x2 – 8x + 16) - (x3 – 3(x2(4)) + 3(x (16)) + (43)
= 2 - x + 4 + x2 – 8x + 16 - x3 + 12x2 - 48x - 64
Pn(x) = - x3 + 13x2 - 57x – 46.
Hence, the variation of the polynomial is same in all those derivatives for the same or different values.
Examples for polynomial variation
4q3 – 10q2 + 5q = 2q × 2q2 – 2q × 5q + 2q × 5/2
= 2q ( 2q2 -5q + 5/2)
Hence, 4q3 – 10q2 + 5q = 2q ( 2q2 -5q + 5/2) , so that
`(4q^3-10q^2+5q)/(2q)` = `(2q( 2q^2-5q+5/2))/(2q)`
= 2q2 -5q + 5/2
In these example. the variation in polynomials are achieved by the change of degrees that acquire the high level of degrees. if the polynomials variation occurs in the division operation the separation of the same terms happen in both of the sides.
In the first step, the separation of degree occurs from 4q3 to 2q × 2q2happens.
In the second step of this solution, the same term 2q is used divide the terms present at both of the terms.
A polynomial that contains only one variable,we take it as x, is known as a polynomial in the variable x.
Thus, 5x2 + 13x – 9 is a polynomial in the variable x.
y3 + 7y – 19 is a polynomial in the variable y.
It is customary to write the terms of a polynomial in decreasing order of the exponents of the variables. This is called the standard form of a polynomial. The term that does not contain the variable is written at the end. It is known as the constant term.
Usage of the polynomial variation:
Solve the f(x) = `(1)/(x-3)` , x = 4. Find the degree of polynomial up to three.
Solution:
f(x) = (1) /( x - 3) f (4) = 2
f ‘(x) = (-1) /(( x - 3)^2) f ‘(4) = -1
f ‘’(x) = (2) /(( x - 4)^3) f ‘’(4) = 2
To find the Taylor polynomial:
General form of the Taylor polynomials is,
Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n
Here, we calculate up to third degree of Taylor polynomial.
Pn(x) = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!) + f’’’ (4) ((x – a)3/ 3!)
= 2 + - (1) (x - 4) + 2((x – 4)2 / 2!) + (-6) (x – 4)3/ 6
= 2 - x + 4 + (x2 – 8x + 16) - (x3 – 3(x2(4)) + 3(x (16)) + (43)
= 2 - x + 4 + x2 – 8x + 16 - x3 + 12x2 - 48x - 64
Pn(x) = - x3 + 13x2 - 57x – 46.
Hence, the variation of the polynomial is same in all those derivatives for the same or different values.
Examples for polynomial variation
- Divide 4q3 – 10q2 + 5q by 2q.
4q3 – 10q2 + 5q = 2q × 2q2 – 2q × 5q + 2q × 5/2
= 2q ( 2q2 -5q + 5/2)
Hence, 4q3 – 10q2 + 5q = 2q ( 2q2 -5q + 5/2) , so that
`(4q^3-10q^2+5q)/(2q)` = `(2q( 2q^2-5q+5/2))/(2q)`
= 2q2 -5q + 5/2
In these example. the variation in polynomials are achieved by the change of degrees that acquire the high level of degrees. if the polynomials variation occurs in the division operation the separation of the same terms happen in both of the sides.
In the first step, the separation of degree occurs from 4q3 to 2q × 2q2happens.
In the second step of this solution, the same term 2q is used divide the terms present at both of the terms.