M10B2011Sampp05Soln

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M10B(Stream A) : Functions of Real Variables

University of the West IndiesDepartment of Mathematics

Problem Paper #05 Solutions

Lecturer: Sam McDaniel

Question 1

(a) Let f : R ! R be such that f(x) = 1=x for all x 6= 0:Find a 4th degree Taylor series polynomial approximation for f near

a = 1: Use your approximation to estimate the value of 1=0:9:(b) Let f : [1; 1) ! R be such that f(x) = px + 1 .

Find a fourth-order estimate for y = f(x) f(0) in powers of x =x 0 = x:

(c) Find the Taylor series expansion for f(x) =sin x

1 x in powers ofx up toand including the term in x4:

Solution 1

(a) f(x) = 1=x for all x 6= 0:The 4th degree Taylor series polynomial approximation for f near x =a is

f(x) = f(a)+f0(a)(xa)+f2(a)

2!(xa)2+f

3(a)

3!(xa)3+f

4(a)

4!(xa)4+

Now

f0(x) = 1x2

; f2(x) =2

x3; f3(x) = 6

x4; f4(x) =

24

x5

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M10B Problem Paper 05 Solutions

and so

f(1) = 1; f0(1) = 1; f2(1) = 2; f3(1) = 6; f4(1) = 24The 4th degree Taylor series polynomial approximation for f near x =

1 is therefore

f(x) = 1 (x + 1) + 22!

(x + 1)2 +63!

(x + 1)3 +24

4!(x + 1)4 +

= 1 (x + 1) (x + 1)2 (x + 1)3 (x + 1)4 + Put x = 0:9 to get10:9

1 (0:9 + 1) (0:9 + 1)2 (0:9 + 1)3 (0:9 + 1)4

= 1 (0:1) (0:1)2 (0:1)3 (0:1)4= 1:1111

(b) f(x) = px + 1 .With y = f(x) f(0) and x = x 0 = x: The Taylor seriesexpansion is

y = f0(0)x +f2(0)

2!x2 +

f3(0)

3!x3 +

f4(0)

4!x4 +

= f0(0)x +f2(0)

2!x2 +

f3(0)

3!x3 +

f4(0)

4!x4 +

Now,

f0(x) = 12

(x + 1)12

f2(x) = 12 12

(x + 1)3

2

f3(x) = 32 1

2 12

(x + 1)5

2

f4(x) = 52 3

2 1

2 12

(x + 1)7

2

resulting in

f0(0) = 12

; f2(0) = 14

; f3(x) = 38

; f4(0) = 1516

and therefore,

y = 12

x 18

x2 + 116

x3 5128

x4 +

(c) To nd the Taylor series expansion for f(x) =sin x

1 x in powers of xup to and including the term in x4; Ill simply write down the series

expansions ofsin x and1

1 x before multiplying them to get the result.

Well most of us know that

sin x = x

1

6x3 + 1

120x5 +

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and that1

1 x = 1 + x + x2 + x3 + x4 +

and so

sin x

1 x = x 16x3

+1

120x5

+ 1 + x + x2

+ x3

+ x4

+ = x + x2 + x3 + x4 1

6x3 1

6x4 +

= x + x2 + 56

x3 + 56

x4 +

Question 2

(a) Show that x7 + x 1 has exactly one real root.(b) Show that x3 3x + 4 = 0 has exactly one real root in [3; 1]:(c) Prove that 2x3

2x2 + 3x = 2 has exactly one root and that this root

lies between 0 and 1:

Solution 2

(a) Let f(x) = x7 + x 1.Since f is a polynomial, f is continuous and dierentiable on R. Con-sider the interval, [0; 1]: f is continuous on [0; 1] and f(0) = 1; f(1) =1 implying that f(0)f(1) < 0: By the Intermediate Value Theorem,9 2 (0; 1) such that f() = 0: So at least one root exists (between 0and 1).

Now assume that f has another root, say : Without loss ofgenerality, let < : We therefore have that f() = 0 = f() andsince f is continuous on [; ] and dierentiable in (; ); by Rollestheorem, 9k 2 (; ) such that f0(k) = 0:But, f0(k) = 7k6 + 1 1 for all k and so there is no 9k 2 (; ) suchthat f0(k) = 0: This is a contradiction to Rolles theorem and thereforethe assumption is incorrect. That is, f has no other root. We concludethat f has exactly one real root (since it has at least one real root) onR.

(b) Let f(x) = x3 3x + 4:f is a polynomial and so f is continuous on [3; 1] with f(3) =27 + 9 + 4 = 14 < 0 and f(1) = 1 + 3 + 4 = 6 > 0: Therefore,f(3)f(1) = 14 6 < 0: By the IMVT, 9 at least one 2 (3; 1)such that f() = 0: So f has at least one real root in the given interval.

Now suppose f has another root in the interval. Withoutloss of generality, let < : f is continuous on [; ] and dierentiablein (; ) with f() = f() = 0: By Rolles Theorem, there existsk 2 (; ) such that f0(k) = 0: Now, f0(x) = 3x2 3 = 3(x2 1) =

3(jxj2

1) = 3(jxj + 1)(jxj 1) > 0 for all jxj > 1: Therefore, here

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are no k 2 (3; 1) and hence no k 2 (; ) such that f0(k) = 0: Thiscontradicts Rolles Theorem and therefore f has no other real root in[3; 1]:

(c) Let f(x) = 2x3 2x2 + 3x 2.

f is a polynomial and so f is continuous on [0; 1] with f(0) = 2 < 0and f(1) = 2 2 + 3 2 = 1 > 0: Therefore, f(0)f(1) = 2 1 < 0:By the IMVT, 9 at least one 2 (0; 1) such that f() = 0: So f hasat least one real root in (0; 1):

Now suppose f has another root . Without loss of generality,let < : f is continuous on [; ] and dierentiable in (; ) withf() = f() = 0: By Rolles Theorem, there exists k 2 (; ) suchthat f0(k) = 0: Now, f0(x) = 6x2 4x + 3 = 6(x 1

3)2 + 7

3 7

3for all

x 2 R: Therefore, there are no k 2 (; ) such that f0(k) = 0: Thiscontradicts Rolles Theorem and therefore f has no other real root on

R: We conclude that f(x) has exactly one root and that this root liesbetween 0 and 1:

Question 3

State Rolles Theorem.

Let f : [2; 3] ! R be such that f(x) = 1 x2=3: Show that f(1) = f(1);but f0(x) is never zero in the interval [1; 1]: Explain why this does notcontradict Rolles Theorem.

Solution 3

Rolles Theorem : Let f be a function that is continuous on [a; b] anddierentiable in (a; b): If f(a) = f(b) then 9c 2 (a; b) such that f0(c) = 0:f : [2; 3] ! R is such that f(x) = 1 x2=3:f(1) = 1 (1)2=3 = 1 [(1)2] 13 = 1 1 13 = 1 1 = 0f(1) = 1 (1)2=3 = 1 [(1)2] 13 = 1 1 13 = 1 1 = 0:Therefore, f(1) = f(1) = 0

Now,f0(x) = 2

3x1=3 = 2

3x1=3;

which does not exist at x = 0: Also, f0(x) = 23x1=3

is never zero. Since f isnot dierentiable at x = 0; it is not dierentiable in (1; 1): Therefore thehypothesis of Rolles Theorem is not satised and so Rolles Theorem is notcontradicted.

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Question 4

Let f : [1; 1] ! R be such that f(x) = x2 11 + x2

: Show that f satises

all the conditions for the application of Rolles Theorem.

Find all c which satisfy the conclusion of the theorem.

Solution 4

Let f : [1; 1] ! R be such that f(x) = x2 11 + x2

:

x2 and1

1 + x2are both continuous on [1; 1] and hence f is continuous on

[1; 1]:

x2 and1

1 + x2are both dierentiable in (1; 1) and hence f is dierentiable

in (1; 1):Also,

f(1) = (1)2 11 + (1)2 = 1

1

2= 1

2

and f(1) = (1)2 11 + (1)2

= 1 12

= 12

:

Hence f(1) = f(1) and so f satises all the conditions for the applicationof Rolles Theorem. By Rolles Theorem, 9c 2 (1; 1) such that f0(c) = 0:That is,

f0(c) = 2c +2c

(1 + c2)2= 0

=2c(1 + c2)2 + 2c

(1 + c2)2= 0

We solve 2c(1 + c2)2 + 2c = 0:

2c[(1 + c2)2 + 1] = 0

resulting in c = 0 only.

Question 5

By considering the function f(t) = tan1 t on [x; 1]; x > 0; prove that

4 1 x

1 + x2< tan1 x 0: f is continuous on [x; 1] anddierentiable in (x; 1) with f0(t) =

1

1 + t2: By the MVT, 9k 2 (x; 1) such

that

f0(k) = f(1) f(x)1 x :

That is,1

1 + k2=

tan1(1) tan1(x)1 x =

4

tan1(x)1 x :

Now, x < k < 1 and therefore, x2 < k2 < 1 ) 1 + x2 < 1 + k2 < 2 implying1

2

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M10B2011Sampp05Soln