Introduction:
In this chapter we discuss about three important theorems Rolle’s Theorem, Mean Value Theorem and Taylor series. Michael Rolle stated a theorem on algebra about the roots of an equation.
This theorem was used in an important result of calculus, now known as Rolle’s Theorem. This is very useful result in calculus and is the basis of mean value theorem. Also our aim in this chapter is to find a polynomial that gives us a good approximation to some functions in terms of power series, also called Taylor series. Such series can be described informally as infinite polynomials i.e. polynomials containing infinitely many terms. This series is used for the expansion of functions. This is very useful for the students in engineering line.
Rolle’s Theorem:
A function f(x) is continuous in the interval [ , ]a b and differentiable on ( , )a b and also ( ) ( )
f a f b then there exists at least one c( , )a b such that f c( )0.
Theorem’s Significance:
Algebraic Significance:
If 𝑓(𝑥) be a polynomial in x and 𝑥 = 𝑎 , 𝑥 = 𝑏 be two roots of the equation 𝑓(𝑥) = 0,then from Rolle’s theorem we find at least one root of the equation 𝑓′(𝑥) = 0 lies between ‘a’ and ‘b’.
Geometrical Significance:
From Rolle’s Theorem we get a tangent parallel to x-axis at the point ‘c’ between ‘a’ and ‘b’.
Problem: Verify Rolle’s Theorem for 𝑓(𝑥) = 𝑥2− 3𝑥 + 2 in the interval (1,2).
Solution: Given function is as follows,
Chapter 05 Various Theorems’
𝑓(𝑥) = 𝑥2 − 3𝑥 + 2 So, f(1) 12 3.1 2 0 f(2)223.2 2 0 f(1) f(2)
The function 𝑓(𝑥) is a polynomial function, so it is continuous and differentiable everywhere.
Differentiating the given function 𝑓(𝑥) with respect to x we get f x( )2x3
Let 𝑐 ∈ (1,2) so that f c( )2c3. From Rolle’s Theorem we have
( ) 0
2 3 0
2 3
3 1.5 (1, 2) 2
f c c c c
It shows that we get one c(1 , 2) such that f c( )0. So the Rolle’s Theorem is verified.
Problem: Verify Rolle’s Theorem for 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) in the interval (1 , 3).
Solution: Given function is as follows,
( )f x (x 1)(x2)(x3) Now, f(1) (1 1)(1 2)(1 3) 0 f(3) (3 1)(3 2)(3 3) 0 f(1) f(3)
We have,
2
3 2 2
3 2
( ) ( 3 2)( 3)
( ) 3 2 3 9 6
( ) 6 11 6
f x x x x
f x x x x x x
f x x x x
The function 𝑓(𝑥) is a polynomial function, so it is continuous and differentiable everywhere.
Differentiating the given function 𝑓(𝑥) with respect to x we get
f x( )3x212x11 Let 𝑐 ∈ (1,3) so that f c( )3c212c11.
From Rolle’s Theorem we have, f c( )0 3c212c 11 0
12 144 4.3.11 12 144 132
2.3 6
12 12
6 c
c
Taking positive sign 𝑐 = 2.57 ∈ (1 , 3) and also taking negative sign 𝑐 = 1.42 ∈ (1 , 3) It shows that we get two c(1 , 3) such that f c( )0. So the Rolle’s Theorem is verified.
Exercise:
(1) Prove the validity of the Rolle’s theorem for the function
2
( ) 1 ( 1)3
f x x in the interval [0,2].
(2) Discuss the applicability of Rolle’s theorem when 𝑓(𝑥) =1+𝑥1 −1−𝑥1 and 𝑎 = −1 , 𝑏 = 1 . (3) Verify Rolle’s Theorem for the functions 𝑓(𝑥) = 𝑥3− 6𝑥2+ 11𝑥 − 6 .
Mean value Theorem:
A function f x( ) is continuous in the interval [ , ]a b and differentiable on ( , )a b then there exists at least one c( , )a b such that ( ) f b( ) f a( )
f c b a
. Geometrical Significance:
The tangent to the curve at ‘c’ is parallel to the chord joining to two ends points.
Problem: Ascertain the validity of Mean value theorem for the function 𝑓(𝑥) = 𝑥(𝑥 − 1)(𝑥 − 2) on the interval [0, 1].
Solution:
Given function is as follows,
f x( )x x( 1)(x2) f(0)0.(0 1)(0 2) 0
(1) 1.(1 1)(1 2) 0
f
Again we have,
( )f x x x( 1)(x2)
( ) ( 2 )( 2)
f x x x x
3 2 2
( ) 2 2
f x x x x x
3 2
( ) 3 2
f x x x x
The function 𝑓(𝑥) is a polynomial function, so it is continuous and differentiable everywhere.
Differentiating the given function 𝑓(𝑥) with respect to x we get f x( )3x26x2
Let 𝑐 ∈ (1,3) so that f c( )3c26c2.
From Mean value Theorem we have, ( ) (1) (0) 0 0 0
1 0 1 0
f f
f c
( ) 0 f c
3c2 6c 2 0
6 36 4.3.2 6 36 24
2.3 6
6 12
6 c c
Taking positive sign c1.57(0,1) and also taking negative sign c0.42(0,1) It shows that we get one c(0 , 1) such that ( ) f b( ) f a( )
f c b a
. So the Mean value Theorem is verified.
Problem: Find the value of c of the mean value theorem 𝑓(𝑏) − 𝑓(𝑎) = 𝑓′(𝑐)(𝑏 − 𝑎) if 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) on interval (0, 4).
Solution: Given function is as follows,
f x( )(x1)(x2)(x3) f(0) (0 1)(0 2)(0 3) 6 f(4) (4 1)(4 2)(4 3) 6 Again we have,
2
3 2
( ) ( 1)( 2)( 3)
( ) ( 3 2)( 3)
( ) 6 11 6
f x x x x
f x x x x
f x x x x
The function 𝑓(𝑥) is a polynomial function, so it is continuous and differentiable everywhere.
Differentiating the given function 𝑓(𝑥) with respect to x we get f x( )3x212x11
Let 𝑐 ∈ (0,4) so that f c( )3c212c11.
From Mean value Theorem we have ( ) (4) (0) 6 ( 6) 3
4 0 4 0
f f
f c
( ) 3 f c
3c2 12c 11 3
3c2 12c 8 0
12 144 4.3.8 12 144 96 6 48
2.3 6 6
6 4 3 6 c
c
Taking positive sign c2.15 (0, 4) and also taking negative sign c 0.15 (0, 4) It shows that we get one c(0 , 4) such that ( ) f b( ) f a( )
f c b a
.so the Mean value Theorem is verified.
Exercise:
(1) Verify Mean value Theorem for the function 𝑓(𝑥) = 𝑥(𝑥 − 1)(𝑥 − 2) on the interval [0,1 ] 2 .
(2) Justify the validity of the mean value theorem for the function 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) in the interval (0 , 4).
(3) Find the value of ‘c’ in the mean value theorem 𝑓(𝑏) − 𝑓(𝑎) = (𝑏 − 𝑎)𝑓′(𝑐) if 𝑓(𝑥) = 𝑥2 , 𝑎 = 0 , 𝑏 = 1 .
Taylor’s series:
Taylor series is an expansion of a function 𝑓(𝑥) about a point 𝑥 = ℎ is given by 𝑓(𝑥 + ℎ) = 𝑓(𝑥) + ℎ𝑓′(𝑥) +ℎ2 !2 𝑓"(𝑥) + ⋯ 𝑡𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦.
Another form of this series for the point 𝑥 = 𝑎 is as 𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎)𝑓′(𝑎) +(𝑥 − 𝑎)2
2 ! 𝑓"(𝑎) + ⋯ 𝑡𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦.
If 𝑎 = 0, the expansion is known as a Maclaurin series.
Conditions: In order to find such a series, some conditions have to be in place:
(1) The function 𝑓(𝑥) has to be differentiable infinitely many times.
(2) The function 𝑓(𝑥) has to be defined in a region near the value 𝑥 = 𝑎 . Problem: Find the Taylor Series for 𝑓(𝑥) = 𝑒−𝑥 about 𝑥 = − 4 .
Solution: Given that 𝑓(𝑥) = 𝑒−𝑥 and 𝑎 = −4 .
Now, takes some derivatives of 𝑓(𝑥) and evaluate them at 𝑥 = − 4.
f x( )ex f '( 4) e4 f x'( ) ex f '( 4) e4 f ''( )x ex f "( 4) e4 f "'( )x ex f '"( 4) e4 Proceeding in this way, we can write
fn( )x ( 1) en x fn( 4) ( 1) en 4 Therefore the Taylor series for f(x) is,
𝑓(𝑥) = 𝑓(− 4) + (𝑥 + 4)𝑓′(− 4) +(𝑥 + 4)2
2 ! 𝑓"(- 4)+(x+4)3
3 ! f"'(− 4) + ⋯ 𝑡𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦.
= (𝑒4) + (𝑥 + 4). (−𝑒4) +(𝑥+4)2 ! 2 . (𝑒4)+(x+4)3
3 ! .(−𝑒4) + ⋯ 𝑡𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦.
40
( 1) .
( 4)
!
n
n n
e x f x n
(Ans)Problem: Find the Taylor Series for 𝑓(𝑥) = 𝑥3 − 20𝑥 + 6 about 𝑥 = 3 . Solution: Given that 𝑓(𝑥) = 𝑥3 − 20𝑥 + 6 and 𝑎 = 3 .
Now, takes some derivatives of 𝑓(𝑥) and evaluate them at 𝑥 = 3.
f x( )x320x6 (3)f 27 f x'( )3x220 f '(3)7 f ''( )x 6x f ''(3) 18 f '''( )x 6 f '''(3)6 Proceeding in this way we get the rest terms are zero.
Therefore the Taylor series for f(x) is,
𝑓(𝑥) = 𝑓(3) + (𝑥 − 3)𝑓′(3) +(𝑥 − 3)2
2 ! 𝑓"(3)+(x-3)3
3 ! f"'(3) + ⋯ 𝑡𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦.
= (−27) + 7. (𝑥 − 3) +(𝑥−3)2 ! 2. 18 +(x-3)3
3 ! .6 + 0 = −27 + 7. (𝑥 − 3) + 9(𝑥 − 3)2 +(x-3)3 ∴ 𝑓(𝑥) = (x-3)3+ 9(𝑥 − 3)2+ 7. (𝑥 − 3) − 27 (Ans)
Exercise:
(1) Find the Tylor series for 𝑓(𝑥) = 𝑒𝑥 about 𝑥 = 2 and hence Maclaurin Series.
(2) Find the Tylor series for 𝑓(𝑥) = 𝑐𝑜𝑠𝑥 about x =3.
(3) Find the Tylor series for 𝑓(𝑥) = 𝑙𝑛𝑥 about = 2 .
