Proposition 1: Any function having a finite derivative in a

Chapter 2

POLYNOMIAL APPROXIMATIONS

Let f be a function having n finite derivatives. We shall give 1i this chapter an estimation of the error in the following polynomial approximation:

n n

a x n

a f a

x a f a x a f a f x

f ( )

! ) ( ...

) ( ! 2

) ( ) ( ! 1

) ( ) ( ) (

) ( 2

" '

 

  

 

 . a

1. PRELIMINARIES

Let I IR be an interval, a I , and f :I IR be a function. We remember that if there exists:

) ( : ) ( : ) ( ) ( lim )

( ) (

lim '

0 _dx a

df a f h

a f h a f a

x a f x f

h h a x a

x  

  



  





,

we call this limit (finite or infinite) the derivative of the function f in a. In the following paragraphs we shall use some known theorems.

Proposition 1: Any function having a finite derivative in a point iscontinuous in that point.

In fact, this Proposition asserts that, in a neighbourhood of a point a, the function f that has a finite derivative f

 

a can be

approximated by a first-degree polynomial:

) a x )( a ( f ) a ( f ) x (

f   '  .

One of our purposes is to generalize this approximation. We shall remember some derivatives for elementary functions: 

 

uv ' _uv1_v_u' _uv_v' _lnu_{, where}

u, v have derivative, u>0



 

chx

2 e e 2

e e shx

x x ' x x '

      



 



 

Rolle’s Theorem. Let f :

 

a,b IR be a Rolle function (i.e. f

is continuous on [a,b] and has finite derivative on (a,b)), such that

f(a) = f(b). Then there exists a î

 

a,b such that f'( )_0.



Fermat’s Theorem. Let f :I IRIRbe a function. If a is an interior point of interval I (i.e.  å_{>0 such that (a-å, a+å)}  _I)

which is a local point of extreme for f and f has a derivative in a, then f'(a)0(i.e. a is a stationary point).

Primitive’s Theorem: If f :

 

a,b IRis a continuous function then:

 

 

_

x

a

dt t f x F b

a

F: , IR, ( ) ( )

is a function for which F’ = f.

Mean Value’s Theorem: If f :

 

a,b IR is a continuous function, there exists a point a î

 

a,b such that

). a b ( ) ( f dx ) x ( f

b

a

  





2. DIFFERENTIABILITY

Definition 1. A function f :I IR is said to be differentiable

in a point a, if there exists a constant AIR and a function IR

 I

:

 continuous in a, with (a)0, such that

) x ( ) a x ( ) a x ( A ) a ( f ) x (

f       

for all xI; the linear function: d_a f :IRIR,d_af(h) Ah

is named the differential of the function f in the point a.

Proof. The necessity. From the hypothesis we know that there is AIR such that

I x ), x ( ) a x ( ) a x ( A ) a ( f ) x (

f         (1) where (a)0and ù _{is a continuous function in a. From relation} (1) we can write that:

a x

a f x f a

f

a

x 

 



) ( ) ( lim ) (

' lim



( )



,

) 1

( xa A x  A

since ù _{is a continuous mapping; therefore f has a finite derivative} in a: A=f’’_(a).

The sufficiency. Suppose now that f has a finite derivative in

a, i.e. the limit:

) ( ) ( ) (

lim f' a

a x

a f x f

a

x  



 ,

exists and it is finite. To prove that f is differentiable in a, let IR

 I

:

 defined by

), a ( f a

x

) a ( f ) x ( f ) x

(  '

  

 if xa, and ù(a) = 0; (2) then:

) ( 0 ) ( ) ( ) ( lim ) (

lim f' a a

a x

a f x f x

a x a

x     

 



 ;

hence ù is a continuous function in a; moreover, from (2) we have: f(x) f(a)(xa)



(x) f(a)



,

hence the function f is differentiable in a and A = f’_(a).

Remarks. 1. The differential in a, of the identity 1_I id_I is denoted by dax, and dax(h)h; obviously daxdbx for all b; therefore, the differential of the independent variable x (i.e. idI) is denoted by dx; with this notation we can write the differential of f in a like this: d_a f  f'(a)dx.

2. If f has finite derivative in every point xI, we say that f is

differentiable on I and

dx

dx df dx f

(where

dx df

f'  is the derivative of f in the current point) is named

the total differential of function f.

3. If f and g are two differentiable functions, then

d



f g



df dg(the linearity of the operator d)

IR  , ,









(if ( ) 0, ) 1

;

' '

2

' '

I x x

g df

g dg f g g

f d

df g dg f g f d

   



 



d(f u) f'du, for all compoundable functions f and u. 4. We remember that, if f :I IR is differentiable, and f’ is differentiable in a, then:

) a ( f : ) a ( dx

f d ) a ( dx

df "

2 2 '

 

     

is named the second order derivative or the second derivative of f; the monomial function of second order:

2 2 " 2

) ( ) ( ,

: d f h f a h

f

d_a IRIR _a 

is called the second order differential in a or the second differential of f in a; since

2 2

h ) h (

dx 

if f has second derivative on I, then: d2f  f"dx2,

where 2 " 2

a f f (a)dx

d  and 2 " 2

a f(h) f (a)h

d  , is named the

second differential of f (on I).

5. Analogously, if there exists f n (a), finite, the n-monomial function:

, ) ( )

( ) ( )

( ,

: n (n) n n n

a n

a f d f h f a dx h f a h

d IRIR   

Example 1. Let f :IRIR, f(x)_x_ex_cos2x. _Then:



e cos2x x e cos2x 2x e sin2x



dx

df  x   x   x

is the total differential of f; for a = 0 we can write the linear function

f

d₀ -the differential of f in 0: d0f (100)dxdx

which is the identity function d0f(h)= h, for all hIR. The second differential can be obtained from the first:

2 2

) 2 cos 4

2 sin 2 2 sin 2

2 sin 2 2 cos 2

cos 2

sin 2 2 cos (

dx x xe

x xe

x e

x xe

x xe

x e

x e

x e

f d

x x

x

x x

x x

x

 



 

 

 

and for a = 0, we obtain a 2-monomial: d02f 2dx2,i.e. d f(h) 2h , h IR,

2 2

0   

which is the second order differential of the function f in the point

a = 0. Analogously:

d03f 9dx3,and ( ) 9 , . 3

3

0 f h  h hIR

d

Exercise 2. Find d02009f , where f(x) = x

2_chx.

Solution. We have Leibniz’ formula:





_







 n

k

k k n k n n

v n C v

u

0

) ( ) ( )

(

, where .

! )! (

!

k k n

n k

n C_nk

        

Here, since

 2009 2 (2009 ) ( )

0 2009 )

2009 )

2009 ( 2

) ( ) ( )

0 ( 2007

, 0 )

( k k

k k k

chx x

C f

k

x 



 _{    }



_|

x=0 =

= 2( ) 2 ( ) 0 2008 2009,

2 2009 2009 '

2008 2009 "

2007

2009  chx C  x chx C x chx x  

C

hence

3.TAYLOR`S FORMULA

Definition 1. Let f :I IRIR, having nth finite derivative in a I. The polynomial:

n n

k n

k

k n

a x a f n a

x a f a x a f a f

a x a f k x

T

) )( ( ! 1 ... ) )( ( ! 2

1 ) )( ( ! 1

1 ) (

) )( ( ! 1 : ) (

) ( 2

" '

0 ) (

 

  

 



 







is named Taylor’s polynomial of n degree, or nth degree Taylor’s polynomialfor the function f in powers of (x-a).

The function

) ( ) ( ) ( ,

:I R x f x T x

R_n IR _n   _n

is called the remainder term of Taylor’s formula, or the nth remainder

(of Taylor_’s formula for a function f in the powers of (x-a)).

The remainder Rn(x) shows us the error that appears when we

replace the exact value of f(x) with the value of Taylor’s polynomial

Tn(x).

If a = 0, Tn(x) is named Mac Laurin`s polynomial, and Rn is

called Mac Laurin`s remainder.

We remark that the Taylor’s polynomial can be written as the following:

). a x ( f d ! n

1 ... ) a x ( f d ! 1

1 ) a ( f ) a x ( f d ! k

1 )

x (

T _an

n

0 k

a k

a

n 



      



In the sequel we will prove a very important formula- Taylor_’s formula- that is a very useful tool in estimating the error that appear by approximation f(x)T_n(x).

Taylor’s Theorem (Taylor_’s formula). Let f :I IRIR, be a function where I is an interval, and suppose that f has continuous derivatives of all orders up to (n+1) in a neighbourhood V of a. Then for all xVand *

IN

p there exists a î between a and x (i.e.



min(a,x),max(a,x)





p

For this we introduce an auxiliary function theF:V IR,

( ) ( ) ( ) .

([a,x] or [x,a]); therefore, according to Rolle`s theorem, there exists

(x ) f ( ). p

! n

1

C n 1 p (n 1)

   

 



Consequently (1) becomes:

f ( )(x ) (x a) ,

! n p

1 ) x ( T ) x (

f (n 1) n 1 p p

n  

 

  





and so the Taylor’s formula is proved.

Remarks. 1.The remainder from the Taylor_’s formula can be expressed in other forms. For instance, since î is a number between x and a, it follows that î can be written as î = a + è(x-a),

where è(0,1), and if p = 1, then we obtain the Cauchy`s remainder:

); 1 , 0 ( , ) 1 ( ) a x ))( a x ( a ( f ! n

1 ) x (

Rn  (n 1)    n 1  n 

 

 



here è depends on n and x.

2. V is a neighbourhood of the point a, hence it results that there is a ä>0, such that [a-ä, a + ä] V; but (n 1)

f  is a continuous function on this closed interval, so it results that, there exists M>0: f(n 1)(x) M, x



a ,a



,

  

   



and the remainder can be evaluated as:

, ;

)! 1 ( )

(  1  



 x a  for x a

n M x

R_n n

this inequality can be used for many purposes: to investigate the behaviour of the nth order remainder, with n fixed, in the neighbourhood of the point a, or to study the remainder when n

tends to infinity, since: 0 ) (

) (

lim 





 n

n

n _{x a}

x R

.

Taylor’s integral formula. Suppose that f :I IRIR, is a function that has continuous derivatives up to the (n+1) order. If

I x ,

a  , then:



_ 

 x

a

) 1 n ( n

n (x t) f (t)dt

! n

1 ) x (

)

remainder in the integral form.

Proof. Using the Newton-Leibniz formula and integrating by parts we obtain consecutively:





proved.

n n n

n x f a x a

n x

R ( ) ( ( ))(1 )

! 1 )

( ( 1)

 

   



 

which is the remainder of Taylor’s formula in powers of (x-a) having the Cauchy`s form.

4. APPROXIMATIONS OF FUNCTIONS (by

polynomials)

Taylor’s formula is a fundamental tool for approximations. By using this formula we can calculate, almost like with a computer, a large class of functions (values of functions). For instance, if we want to compute f(x) with an error less than å > 0, we know that

, ) x ( R ) x ( T ) x (

f  _n  _n

and if T_n(x) , for some nIN, then f(x)T_n(x),

and by that we mean that, when we replace f(x) with Tn(x) the error

is less than å. Moreover, if we know how to calculate

} n ,...., 1 , 0 { k ), a (

f(k)  ,

then Tn(x) can be easily estimated with a computer.

Exercise 3. Find a number, nIN such that:

  

   



3 1 , 3 1 ,

1000 1 )

(x x

T

e n

x

,

Tnbeing the Mac Laurin`s polynomial for f(x)ex;compute 3 e with

three exact digits (decimals).

Solution. According to Mac Laurin`s formula, for all

  

 

3 1 , 3 1

x , there exists a î between x and 0 such that:

, )! 1 ( !

1

0



e n

x k x e

n n

k k

x _

 

 

and

Exercise 4. Approximate 3 _{26, with exactly three decimals.}

Solution. Let 3

1

) (x x

f  . We shall apply the Taylor_’s theorem (with Lagrange_’s remainder), for 3

3

f can be exactly calculated).

We have:

. 962 . 2 27 80

) 1 ( 1 3 1 3 3 1 3 ) 27 26 )( 27 ( ) 27 ( ) 26 (

26 1 ' 3

3

 

         

   



T f f

Mac Laurin`s Formulas for Most Important

Elementary Functions

The function ( ) x.

e x

f  This is a function infinitely differentiable (has finite derivatives of any order) on IR, and

1 ) 0 (

)

(k _

f . Therefore Mac Laurin`s formula with Lagrange_’s remainder is written as the following:





  n

k

n k x

x R n x e

0

), (

! ( ) ( 1)!,

1   



n x e x R

n n



where î is between 0 and x. If x



a,a



,a0, then

  

 

  n

n a e x R

n a

n 0,

)! 1 ( ) (

1

.

This leads to the conclusion that the exponential function can be written as a sum of a series:

,



,



. !

0

a a x n x e

n n

x _



_{  } 



But a is an arbitrary positive number, so this equality is valid on the entire IR.

Example 5. Let us compute the number e, within 10-3. Solution. By taking x = 1, there is 

 

0,1 such that

)! 1 (

3 )!

1 ( ) 1 ( ) 1 (

    



n n

e R

T

e n n



. It is sufficient to solve the inequality:

3000 )! 1 ( 1000

1 )! 1 (

3 _{   }

 n

n .

This relation is fulfilled for n = 6. In conclusion

2.718,

! 6 1 ! 5 1 ! 4 1 ! 3 1 ! 2 1

2     

with an accuracy of 0.001. formula has, in a similar way, the form:

,

Finding Limits with Taylor

’

s Formula

Example 6. Find lim 1



2 3 3 2 6 6ln(1 )



using Taylor’s formula.

Solution. Using the extension of logarithmic function we

therefore

2

), ( !

3 2

1

2 .... ! 4 ! 2 1 2 .... ! 3 2

2 )

(

4 4

4 2 3

x R x

x x x

x x chx

xshx x

f

  

     

 

       

 

    



and

); ( ! 2

... ! 3 ....

! 4 ! 2 1 sin

cos )

(

' 4 4

6 2 4

2 2

2 2

x R x

x x x

x x

x x

x x g

  

    



 _{ }

    



 _{  }

 



according to Taylor_’s formula there exists , '_

 

0,x



 such that:

5 ) 5 ( 4

! 5

) ( )

(x f x

R   and 5

' ) 5 ( '

4

! 5

) ( )

(x g x

R   .

In conclusion, the limit is:

6 1

! 5

) ( 2

1

! 5

) ( !

3 2

1 lim

5 ' ) 5 ( 4

5 ) 5 ( 4

0 

 

  



x g

x

x f

x l

x _



.

Exercise 8. Find



3 4 6 12 12 (ln(1 ) ln )



lim x x2 x3 x4 x5 x x l

x      

 

 .

Solution. In order to apply the Taylor’s formula, we denote

x





 

0,1). (

, 5 12

) ) 5 1 ( 6 5 4 3 2 ( 12 12 6

4 3 1 lim

) 1 ln( 12 12 6

4 3 1 lim

) 1 ln( 12 12 6 4 3 lim

5 6 5

4 3 2 2

3 4 5 0

2 3 4 5 0

5 4 3 2 0

0

 

    

 

      

   

  

   

    



 _{    }



   





y y

y y y y y

y y y y

y y

y y y y

y y

y y y y l

y y y y

5. FINDING THE EXTREMA

We remember that the function f :I IR attains its local maximum (minimum) in a point a, if there is a neighbourhood VVa, such that:

I V x a

f x f a

f x

f( ) ( )0,( ( ) ( )0),   .

According to Fermat`s theorem, if f admits an extreme in a point a and there exists the derivative in this point (i.e. f’_{(a)), then}

this derivative is zero. By definition, a point aI is called a stationary point of the function f, if there exists a derivative of f in that point and f'(a) = 0.

If a function f is defined on the open interval (á, â), and we need to find all its extreme points, then first of all we should find its stationary points, and then we should search among the points where f has no derivative, if such points exist. Stationary points are to be found by solving the equation:

0 ) (

' _

x

f (*)

Of course, not any stationary point of the function f is a local extreme point of f. Condition (*) is necessary for a differentiable function f to have a local extreme in a point x, but not sufficient.

For instance, x=0 is a stationary point for the function ,

, )

( ,

:_{IRIR } 2 1 _IN

n x x f

Theorem 2. Let f :I IR be a function. Suppose that there is aIsuch that:

0 ) ( ....

) ( )

( " ( )

' _{   }

a f a

f a

f n and f(n1)(a)0,nIN_, and also (n1)

f is continuous in a. Then: (a) if n+1 is even then in the point a:

(a1) f has a local maximum if f(n1)(a)0

(a2) f has a local minimum if f(n1)(a)0;

(b) if n+1 is an odd number, then a is not a local extreme for f. Proof. Taylor’s theorem implies the fact that there is a number î between a and x such that:

) ( )!

1 (

) ( ) ( )

( ( 1)

1

  

  

 n

n

f n

a x a f x

f . (1) Since (n1)

f is a continuous function in a, and f(n1)(a)0, it follows that there is a neighbourhood of the point a, such that:

0 ) ( when , ,

0 )

( ( 1)

) 1

(  _{   }  _

a f I

V x x

f n n ₍₂₎

or:

0 ) ( when , ,

0 )

( ( 1)

) 1

(  _{   }  _

a f I

V x x

f n n . (3)

Case (a): n+1 is even:

(a1) If f(n1)(a)0, then from (1) and (2) we have that:

I V x a

f x

f( ) ( )0,   -{a},

since (  )n1 0

a

x , therefore a is a local minimum point.

(a2) If f(n1)(a)0, then from (1) and (3) we have that:

I V x a

f x

f( ) ( )0,   ,

therefore a is a local maximum point. Case (b): n+1 is odd:

Here 1

) (  n

a

x has different signs for x < a and for x> a, and from (1) it follows that f(x) f(a) has also different signs in any neighbourhood of the point a, i.e. a is not a local extreme.

x x x

f( )sin  ,

! 3 sin

) (

3 x x x x

g    ,

x x

x

h( )4 2 32cos2 .

Solution.1. For solving the equation (*): f'(x)0 we have

ZI k k x x

x

f'( )cos 10 _k 2 ,  .

But

"( )_sin _0 k

k x

x

f and (3)( )_cos _1

k

k x

x

f .

It results that n+1 = 3 is not even; therefore all stationary points for f

are not local extremes. 2. We have:

IR

        

 

 x x g x x x f x x

x

g ( ) sin ( ) 0,

2 1 cos )

( "

2

' _.

It results that g' is a monotone increasing function, therefore a = 0 is a unique stationary point. Moreover

'(0)_ "(0)_ (3)(0)_ (4)(0)_0

g g

g

g and (5)(0)_1_0

g .

In conclusion, n+1= 5 is an odd number and g has no local extremes.

3. Since

), 2 ( 4 2 sin 4 8 ) (

'

x f x x

x

h   

x=0 is the unique stationary point, '(0)_ "(0)_ (3)(0)_0

h h

h and (4)(0)_32_0

g ,

so it results that n+1 =4, is even and a = 0 is a local minimum point, and:

hmin=h(0)=3+2=5.

6. ALGEBRAIC APPLICATION

Let Pnbe the linear space of real polynomial functions having

the degree less or equal to nIN, over the field IR, Bc= {1,x,…,xn}

be the canonical basis, and B={1,x-a…_(x-a) n_}(a _{0) another basis}

for Pn. If

n

P

f  , n

nx

a x

a a x

is a vector written is the basis Bc, since f(n1)(x)0,xIR, then: Using Newton’s binomial formula:

i formula) is:







Exercise 11. Find the inverse of the matrix:

   

 

   

 





1 0 0 0

3 1 0 0

3 2 1 0

1 1 1 1

T 1 ;

(since: _1_ 1( _1)_ 2( _1)2 _...).

x C x

C

xi _{i i}

7. SOLVED PROBLEMS

Exercise 12. Expand the polynomial function:

2 5 5

2 )

(x x x x

f    ,

on powers of x1 and x1.

Solution. Since f(6)(x)0 for all xIR we obtain the remainder in Taylor's formula R5(x)=0, for all xIR; therefore:

k k

k

a x k

a f x

T x

f ( )

! ) ( )

( ) (

5

0 ) (

5  







where a1, and,in the second case a1. Then: 0

) 1 ( 

f ; f(1)1122;

x x x x

f

'( )_5 4 _4 3 _4 _,

5 ) 1 (

' 

f

,

_f

'(1)5; 4

12 20

)

( 3 2

''

 

 x x

x

f

,

_f

''(1)28;

_f

''(1)12;

x x

x

f

'''( )_60 2 _24 _,

84 ) 1 (

'' '



f

;

_f

'''(1)36; 24

120 ) (

) 4 (



 x

x

f

,

_f

(4)(1)144;

_f

(4)(1)96; !

5 ) (

) 5 (

 x

f

, hence







1



,

1 6 ) 1 ( 14 ) 1 ( 14 ) 1 ( 5 ) 1 ( !

) 1 ( )

(

5 5

0

4 3

2 )

(

 

    

    







x

x x

x x

x k f x

f

k

and

. ) 1 ( ) 1 ( 4

) 1 ( 6 ) 1 ( 6 ) 1 ( 5 2 ) 1 ( !

) 1 ( )

(

5 4

3 2

5

0 ) (

   

          







x x

x x

x x

k f x

f k

k k

Exercise 13. Let x

e x x

f( ) 2  . Compute: (a) ( )( )

x f n _{, for}

nIN*. (b) d₀f , d0f(x), d f

2

0 , ( )

2 0 f x

d , d₀3f , 3 ( )

0 f x

d .

(c) d₁2008f , d12008f(1).

Solution. (a) Using Leibnitz’ formula, for n2, we have that:



     







 

 



 



 

  









2 1

2

2

) ( 2 )

( 2 ) ( 0

) (

) 1 )( 1 ( 2

) 1 ( ) 1 (

) ( ) 1 ( )

( ) ( )

(

n n

n x

n

n k

k x k n k n k

k n x n

k k n n

n n nx x

e

x e C

x e

C x

f

=(_1) 



2 _2 _ ( _1)



n n nx x

e x

n _,

and we observe that this formula is true for nIN. (b) Generally n n n

a f f a dx

d _ ( )( ) _and n n n

a f x f a x

d ( ) ( ) , for all nIN.

Here:

d0f  f'(o)dx0 , d0f(x)0; 2

2 2

0 f f ''(0)dx 2dx

d   , d₀2f(x)2x2;

3 3

3

0 f f' ''(0)dx 6dx

d   , 3 3

0 f(x) 6x

d  .

(c) 2008 2008 2008

1 f e (1 2 2008 2008 2007)dx

d       ,

2008 2008

1 (1) 4026041





 e

f

d .

Exercise 14. Using Mac Laurin's formula compute:



x chx



x l

x 2 2

12

lim 2

4

0  



)

We observe that n2 verifies this inequality; therefore

Exercise 16. Prove that

n

Solution. The exponential function can be written as a sum of series:

),

whence:

!

and the inequality is proved.

5

0 10

1 ! 1 ! 1

0 

  





 k n n

e

n

k

,

and we obtain for 2.71828.

! 1 )

1 ( :

8

8

0

8





 

 

k k

T e n

Exercise 17. Write the vector

 

x x

x x x

f( )13 3 2  3IR₃

in the base _



1, _2,( _2)2,( _2)3



.

x x

x B

Solution. Let a2 in Taylor's formula. Then there exists an  between 2 and x such that the remainder would be:

0 ) 2 ( ! 4

) ( )

( 4

) 4 (

4  x 

f x

R  ,

for all xIR; hence:

. ) 2 ( !

) 2 ( )

(

3

0 )

( k

k k

x k f x

f





 

But:

! 3 '' '

, 6 ) 2 ( '' , 6 6 ) ( ''

; 3 ) 2 ( ' , 3 6 3 ) ( ' ;

3 2

) 2 (



 

 

 

  

f

f x x

f

f x x x

f f

and

In conclusion: _3_3( _2)_3( _2)2 _( _2)3.

x x

x f

Exercise 18. Let f: I IR_→IR, be a function where I is an interval and suppose that there exists ( 1)( )

a

f n IR where aI.

Show that:

), ( )!

1 (

1 )

(

) ( ) (

lim ( 1)

1 f a

m a

x

x T x

f m

m m a

x

 

   



_

_

n m 1,2,...,



where Tm is Taylor's polynomial of degree m associated to f in the

point a.

). ( )!

1 (

1 )

( 2 ... ) 1 (

) ( )

( lim )

(

) ( ) (

lim ( 1)

) ( )

(

1 f a

m a x m m

a f x f a

x

x T x

f m

m m

a x m m a

x

 



     

 

 

Exercise 19. Prove the inequality: ,

8 2 1 1

2 x x

x    for all x>0.

Solution. Let f :



1,



R, f(x) x1. According to the Taylor's theorem :

 

0,x 

 so that 2

2 ) ( '' )

0 ( )

(x f x f x

f     ,

hence:



1



.

8 2 1

1 2

3 2



 

 

 x x 

x

Therefore

, 8 2 1 1

2 x x

x    for all x<0.

Exercise 20. Using Taylor's formula prove that:

2 5

cos 4 2 1 lim

2

4 2

0 2 

   

 

x e

x x chx

x

x .

Solution. Using the well known formulas for elementary functions it follows that there exists c and c’ _{between 0 and x such}

that:

, 5 4

2

! 5 ! 4 ! 2

1 x x shc x

chx    

.

Therefore:

 

   

   

 

  

 )

315 7 15 2 3 ( 315

17 10

) ! 3 ( 5 15

2 3 5 7

7 3

3 4

5 x x x

x x x

x x x



) 10 ) 3 ( 5 ( ! 5

1 ) 9 3 ) 15 2 3 ( 3 ( ! 3

1 3

3 4 5 6

5 3 2 3

x x

x x x

x x

x x

x       

 =

=             



5 3

7

0 ₁₅)

2 ! 5 1 ! 3 1 15

2 ! 3

1 ! 5 1 ( ) ! 3

1 3 1 3 1 ! 3 1 ( ) 1 1 [( 1

lim x x x

x

x

 

          

 ) ]

! 4 3

1 6 3

1 15

1 315

17 6 3

2 36

1 15

1 ! 7

1 ! 7 1 315

17

( x7

. 30

1 5 3 2

6 5 3 2

7 5

3 2

1 6

5 3

11 36

3 ! 5 3

2

2 2 2

2 2

2 _  _{ }  

       

 _

8.EXERCISES

Exercise 1. Write the following polynomial functions after the powers of x+1 and x+2:

a) ( )_ 3 _ 2 _{ }1

x x x x f

b) ( )_ 4 _ 3 _ 2 _5

x x x x f

c) f(x) x5 2x3 3x

Exercise 2. Write the following polynomial functions after the powers of x-1 and y+1:

a) f(x,y) x2 xyy2 x5 b) f(x,y) x3 x2yxyy2 x1 c) f(x,y) x4 y3 xy2 xy

Exercise 3. Using Taylor's formula, show:

Exercise 4. Find the extrema of the following functions: (a) f(x)cosxsinx,x

  



2 , 2

 

(b) f(x)sin2xx,

  

 

2 , 2

 

x

(c) f(x)xtanx

(d) f(x)sin3x3sinx

(e) f(x)2xarctanx

(f) f(x)sinxcos2 x

(g) 

  

      

2 1 , 2 1 ,

2 6 )

( 11 6

x x x x f

(h) ( )_ 2 _2cos _3

x x

x

f .

Answer. (a) 2

4

max 

      f 

f ;

(b) 

             

6 ,

6 min

max

 

f f f

f ;

(c) there is neither maximum nor minimum;

(d) 

     

2

min



f

f , 

     

2 3

max



f

f ; (e) no extrema;

(f) 

     

2

min



f

f , _

  

   

3 2 arccos

max f

f , ))

3 2 (arccos(

max  f 

f ;

(g) fmin  f(0)2; (h) fmin  f(0)5.

Exercise 5. Find the image of the functions: (a) f :



2,2



IR, ( )_3 4 _6 2 _1

x x x

f

(b) f :



1,5



2 3 1

3 1 ) (

,  3 2  

IR f x x x x

(c)

 

1 1 )

( , 4

, 0 :

   

x x x f

f IR

(d) f  f x  xx

  

 , ( ) sin2

2 , 2

(e) 2 4

12 cos 24 ) ( ,

: f x x x x

f IRIR   

(f)





3 arccos

) ( , 1

, 1 :

3 x x x x

f

f  IR    .

Answer. (a)



25,2



; (b)

  



3 23 , 3 13

; (c)

   

5 3 ,

1 ; (d)

  



2 , 2

 

;

(e)



,24



; (f)

  

 _

6 5 , 6 7

 .

Exercise 6. Prove the inequalities: (a)      , 



0,



! 5 ! 3 sin

! 3

5 3 3

x x x x x x

x ;

(b)       , 



,



3 2 )

1 ln( 2

3 2 2

o x x x x x x

x ;

(c) 1-x  x  x  x ,xIR !

4 2 1 cos 2

4 2 2

.

Exercise 7. Find the extrema of the following functions:

(a) f  f x  x x x

3 arctan )

( , :

3

IR

IR ;

(b) , ( ) 2 8

3 1 , 3 1

: _{ } 6 _ 3 _

   



  f x x x

f IR ;

(c) f :



1,1



IR, f(x)x10 2x5 13. Answer: (a),(b),(c) No extrema.

Exercise 8. Find the extrema of the following functions:

(a) f :

 

0,1 IR, f x  x x x

3 arctan )

(

3

; (b) f :

 

0,1 IR, ( )_2 6 _ 3 _8

x x x

f ;

(c) f :

 

o,1 IR, ( )_ 10 _2 5 _13

x x x

Answer. (a) fmin  f (0)0,

3 2 4 ) 1 (

max   



f

f ;

(b)

8 63 4 1

3

min 

      f

f , fmax  f(0)8, fmax  f(1)9;

(c) fmax  f(0)13, fmin  f(1)12.

Exercise 9. Using Taylor's formula prove that: (a) lim 3₄



2 cos( 2 )



1

0  



 _x e x

x

x ;

(b)



12ln 3 16 36 48 25



1

) 1 ( 12

5

lim ₅ 4 3 2

1       

 _x x x x x x

x ;

(c)



       

 ( 6 ) 6 ( 5 ) 15 ( 4 ) 20 ( 3 )

1 lim ₆

0_h f x h f x h f x h f x h

h



( ), )

( ) ( 6 ) 2 (

15 (6)

x f x f h x f h x

f     



where f :IRIR and f (6) on IR.

(d) lim 2₃



tan sin



1

0  

 _x x x

x ;

(e) lim2 2ln 1 _1

  

 





 _x

x x x

x ;

(f)



ln(1 2 ) 2 sin2



1 4

1

lim ₃ 2

0    

 _x x x x

x ;

(g) 1

) 2 2

( 6

cos )

sin( lim

2 2

0   



 _{xshx chx}

x x x

x ;

(h) sin 1

3 2

! 7 lim

3 7

0 _ 

 

 

 

 x

x shx x

x .

Exercise 10*. Let f :IRIR having the second derivative on IR, such that f(x)13, f'(x) 13 for all xIR. Prove that

26 ) ( ' x 