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:
f ( x )  f (a ) 

f ' (a )
f " (a )
f ( n ) (a )
( x  a) 
( x  a ) 2  ...... 
( x  a)n . a
1!
2!
n!

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

f ( x)  f (a )
f (a  h )  f (a )
lim
lim
: f ' ( a ) :
(a ) ,

x a
h0
dx
xa
h
x a h
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 is continuous 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:

f ( x )  f ( a )  f ' ( a )( x  a ) .
One of our purposes is to generalize this approximation.
We shall remember some derivatives for elementary functions:
 u v   u v 1  v  u '  u v  v'  ln u , where u, v have derivative, u>0
'

 e x  e x
 shx   
2

'
 chx   shx .
'

'

 ex  ex
 
 chx
2



89

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  IR be 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:
function then:

If

f : a, b  IR is a continuous

x

F : a, b  IR, F ( x )   f (t )dt
a

’

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
b

 f ( x )dx 

f (  )  ( b  a ).

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
 : I  IR continuous in a, with ( a )  0 , such that

f ( x )  f ( a )  A  ( x  a )  ( x  a )  ( x )
for all x  I ; the linear function:
d a f : IR  IR, d a f ( h )  A  h
is named the differential of the function f in the point a.
Theorem 1. For a function f : I  IR ( I  IR an interval) to
be differentiable in a point a  I it is necessary and sufficient that it
has a finite derivative at this point; in this case A  f ' ( a ) .

90

Proof. The necessity. From the hypothesis we know that
there is A  IR such that
f ( x )  f ( a )  A  ( x  a )  ( x  a )   ( x ), x  I
(1)
where ( a )  0 and ù is a continuous function in a. From relation
(1) we can write that:
f ( x )  f (a )
f ' ( a )  lim
 limA   ( x )  A,
x a

(1) x  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:
f ( x )  f (a )
lim
 f ' (a ) ,
x a
xa
exists and it is finite. To prove that f is differentiable in a, let
 : I  IR defined by
f( x) f(a)
( x ) 
 f ' ( a ), if x  a, and ù(a) = 0;
(2)
xa
then:
f ( x)  f (a )

lim  ( x )  lim
 f ' (a )  0   ( a ) ;
x a
x a
xa
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 1I  id I
denoted by d a x, and d a x( h )  h ; obviously d a x  d b x for all
therefore, the differential of the independent variable x (i.e. id I )
denoted by dx ; with this notation we can write the differential of f
a like this: d a f  f ' ( a )dx .

is
b;
is
in

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

differentiable on I and
df
df  f '  dx 
 dx ,
dx

91

df
is the derivative of f in the current point) is named
dx
the total differential of function f.

(where f ' 

3. If f and g are two differentiable functions, then
d f   g   df   dg (the linearity of the operator d )
 ,   IR ,
d  f  g   f ' dg  g ' df ;

d



1
f
 2 f ' dg  g ' df
g g



(if g ( x )  0, x  I )

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:
'

d2 f
 df 

( a ) : f " ( a )
  (a)
2
dx
dx
 
is named the second order derivative or the second derivative of f;

the monomial function of second order:
d a2 f : IR  IR , d a2 f ( h )  f " ( a )h 2
is called the second order differential in a or the second differential
of f in a; since
dx 2 ( h )  h 2
if f has second derivative on I, then:
d 2 f  f " dx 2 ,
where d a2 f  f " ( a )dx 2 and d a2 f ( h )  f " ( a )h 2 , is named the
second differential of f (on I).
5. Analogously, if there exists f n  (a ) , finite, the n-monomial
function:
d an f : IR  IR, d an f ( h )  f ( n ) (a )  dx n (h )  f n ( a )h n ,

is called the nth order differential in the point a.

92

Example 1. Let f : IR  IR, f ( x )  x  e x  cos 2 x. Then:
df  e x cos 2 x  x  e x cos 2 x  2 x  e x sin 2 x dx
is the total differential of f; for a = 0 we can write the linear function
d 0 f -the differential of f in 0:
d 0 f  ( 1  0  0 )dx  dx
which is the identity function d 0 f ( h ) = h, for all h  IR . The second
differential can be obtained from the first:
d 2 f  ( e x cos 2 x  2e x sin 2 x  e x cos 2 x  xe x cos 2 x  2 xe x sin 2 x 
 2e x sin 2 x  2 xe x sin 2 x  4 xe x cos 2 x )dx 2
and for a = 0, we obtain a 2-monomial:
d 02 f  2 dx 2 ,i .e. d 02 f ( h )  2 h 2 ,h  IR ,
which is the second order differential of the function f in the point
a = 0. Analogously:
d 03 f  9 dx 3 , and d 03 f ( h )  9h 3 , h  IR.

Exercise 2. Find d 02009 f , where f(x) = x2chx.
Solution. We have Leibniz’ formula:
n

u  v ( n )   C nk n ( n k ) v ( k ) ,
k 0

n
n!
where C nk    
.
 k  ( n  k )! k!

Here, since
( x 2 ) ( 2009 k )  0, k  2007  f 

2009 )

2009

C

( 0) 

k
2009

( x 2 ) ( 2009 k ) ( chx ) ( k ) |x=0 =

k 0

=C

2007
2009

 2(chx )  C
"

2008
2009

2009 2
 2 x (chx )  C2009
x chx
'

x 0

 2008  2009,

hence
d 02009 f  2008  2009 dx 2009 , and d 02009 f ( h )  2008  2009  h 2009 .

93

3.TAYLOR`S FORMULA
Definition 1. Let f : I  IR  IR, having nth finite derivative
in a  I . The polynomial:
n

Tn ( x ) :

1

 k! f

(k )

( a )( x  a ) k 

k 0

 f (a ) 

1 '
1
1
f ( a )( x  a )  f " ( a )( x  a ) 2  .....  f ( n ) ( a )( x  a ) n
1!
2!
n!

is named Taylor’s polynomial of n degree, or nth degree Taylor’s
polynomial for the function f in powers of (x-a).
The function
Rn : I  IR, Rn ( x )  f ( x )  Tn ( x )
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:
n

1 k
1
1
d a f ( x  a )  f ( a )  d a f ( x  a )  ......  d an f ( x  a ).
1!
n!
k 0 k !

Tn ( x )  

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 )  Tn ( 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  V and p  IN * there exists a î between a and x (i.e.
  min( a , x ), max( a , x ) such that:

94

k

1 (k)
f ( n 1 ) (  )
f ( x )   f ( a )( x  a ) 
( x   ) n  1 p ( x  a ) p
p  n!
k 0 k !
or, equivalent
n

(*);

f ( x )  Tn ( x )  Rn ( x ) , where
f ( n 1 ) (  )
( x   ) n  1 p ( x  a ) p .
p  n!
The expression (*) is named Taylor’s formula (Mac Laurin`s
formula for a=0). If p = n+1 we obtain a so called remainder of
Lagrange:
f ( n 1 ) (  )
1
Rn ( x ) 
( x  a )n 1 
d n 1 f ( x  a ).
( n  1 )!
( n  1 )!
Rn ( x ) 

Proof. We want to show that there is constant C  C ( a , x , p )
such that:
1
1
( x  a ) f ' ( a )  ...... ( x  a ) n f ( n ) ( a )  C( x  a ) p . (1)
1!
n!
For this we introduce an auxiliary function the F : V  IR,
1
1
F (t )  f (t )  ( x  t ) f ' (t )  ...  ( x  t ) n f ( n ) (t )  C ( x  t ) p .
(2)
1!
n!
We remark that F(x) = f(x) (from (2) ) and F(a) = f(x) (from (2)
and (1)), hence, it results that F(x) = F(a); but F is a Rolle function
([a,x] or [x,a]); therefore, according to Rolle`s theorem, there exists
a î between a and x such that:
(3)
F' ( )  0 .
But:
1
1
F ' ( t )  f ' ( t )   f ' ( t )  ( x  t ) f " ( t )   2 f " ( t )  ( x  t ) 2 f ( 3) ( t ) 
2!
1!
1
 ...   n ( x  t ) n 1 f ( n ) (t )  ( x  t ) n f ( n 1) (t )  p  C ( x  t ) p 1
n!
hence:
1
F ' ( t )  ( x  t ) n f ( n 1 ) ( t )  p  C ( x  t ) p 1 ,
n!
and from (3) we obtain that
f( x)  f(a)













95

1
( x   ) n 1 p f ( n 1 ) (  ).
n! p
Consequently (1) becomes:
1
f ( x )  Tn ( x ) 
f ( n 1 ) (  )( x   ) n 1 p ( x  a ) p ,
p  n!
and so the Taylor’s formula is proved.
C

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 ( n 1 )
Rn ( x ) 
f
( a   ( x  a ))( x  a ) n 1 ( 1   ) n ,  ( 0 ,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 f ( n 1 ) 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:
M
n 1
Rn ( x ) 
xa
, for x  a   ;
( n  1)!
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:
R ( x)
lim n n  0 .
n  ( x  a )
Taylor’s integral formula. Suppose that f : I  IR  IR , is
a function that has continuous derivatives up to the (n+1) order. If
a , x  I , then:
1 x
Rn ( x )   ( x  t ) n f ( n 1 ) ( t )dt ,
(1)
n! a
and:

96

f ( k )( a )
( x  a ) k R n ( x ) .
(2)
k!
k 0
The formula (1) - (2) is called Taylor’s formula with the
remainder in the integral form.
n

f(x) 

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

x

f ( x )  f ( a )   f ' (t )dt  f ( a )   (t  x ) '  f ' (t )dt 
a

a

x

 f ( a )  (t  x ) f (t ) |   ( x  t ) f " (t )dt 
'

x
a

a

1 x
(( x  t ) 2 ) ' f " (t )d t 
2! a
1 x
1
 f ( a )  ( x  a ) f ' ( x )  ( x  t ) 2 f " (t ) |ax   (( x  t ) 3 ) ' f "' (t )dt 
3! a
2!
1
 f (a )  ( x  a ) f ' ( x)  ( x  a ) 2 f " (a ) 
2!
x
1
 ( x  a ) 3 f "' (t ) |ax   ( x  t ) 3 f ( 4 ) (t )dt 
a
3!
1
1
 f ( a )  ( x  a ) f ' ( x )  ( x  a ) 2 f " ( a )  ( x  a ) 3 f "' ( a ) 
3!
2!
1 x
  ( x  t ) 3 f ( 4 ) (t )dt 
3! a
n
1 x
1
 ...   ( x  a ) k f ( k ) ( a )   ( x  t ) n f ( n 1) (t )dt ,
n! a
k  0 k!
and the Taylor’s formula with the remainder in the integral form is
proved.
 f (a )  ( x  a ) f ' ( x) 

Remark. Applying the mean value theorem to the remainder
(1) it results that there is a î between a and x such that:
1
Rn ( x )  ( x   ) n f ( n 1 ) (  )( x  a ),
n!
or by putting
  a   ( x  a ),  0 ,1 ,
we will obtain:

97

1
( x   ) n f ( n 1) ( a   ( x  a ))(1   ) n
n!
which is the remainder of Taylor’s formula in powers of (x-a) having
the Cauchy`s form.
Rn ( x ) 

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
f ( x )  Tn ( x )  Rn ( x ) ,
and if Tn ( x )   , for some n  IN , then
f ( x )  Tn ( 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
f ( k ) ( a ), k  { 0 ,1,...., n } ,
then Tn(x) can be easily estimated with a computer.
Exercise 3. Find a number, n  IN such that:
1
 1 1
e x  Tn ( x ) 
, x    ,  ,
1000
 3 3
Tn being the Mac Laurin`s polynomial for f ( x)  e x ; compute
three exact digits (decimals).
Solution. According to Mac Laurin`s formula, for all
 1 1
x    ,  , there exists a î between x and 0 such that:
 3 3
n
xk
x n 1
ex  

 e ,
(n  1)!
k  0 k!

98

3

e with

and
Rn ( x ) 

1
1
1
3
1
.
 n 1 3 e  n 1
 
n
( n  1)! 3
3 ( n  1)! 2 2  3 ( n  1)!

R3 ( x ) 

1
1
1
 1 1


, x    , .
2  27  24 1296 1000
 3 3

Then:

So finally
3

1
1
1
1
4
10
113
e  T3 ( )  1   2
 3
 

 1.395
3
3 3  2! 3  3! 3 27  6 81

Exercise 4. Approximate

3

26 , with exactly three decimals.

1
3

Solution. Let f ( x)  x . We shall apply the Taylor’s theorem
(with Lagrange’s remainder), for a  27  33 (because
f (27), f ' ( 27), f " ( 27),... can be exactly calculated).
We have:
1
1
1 3 1 "
1  1  3 2
'
f ( x )  x , f ( x )    1 x ,...
3
33 
1

11 
1
 k
.., f ( x )    1          k  1 x 3 .
33 
3

For x = 26 and a = 27, using the Taylor’s theorem it follows that,
there exists   26,27  , such that:
(k )

f ( n 1) ( )
(1) n 1 , and f ( x)  Tn ( x)  Rn ( x);
Rn ( x ) 
(n  1)!
moreover:
1
1  1 
1
1  3 ( n 1)
1



Rn ( x )  1   2         n  
3  3 
3
3
( n  1)!


1  1
1
3

 1         n   n 1
3  3
3  26 ( n  1)!

We remark that:
1 1 3
1
1


,
R1 ( 26)  1   2
3  3  26  2 3  676 1000
and finally we obtain that:

99

.

1 1 
26  T1 ( 26)  f ( 27)  f ' ( 27)( 26  27)  3   33   1( 1) 
3 3 
80

 2.962.
27

3

Mac Laurin`s Formulas for Most Important
Elementary Functions
The function f ( x)  e x . This is a function infinitely
differentiable (has finite derivatives of any order) on IR, and
f ( k ) (0)  1 . Therefore Mac Laurin`s formula with Lagrange’s
remainder is written as the following:
n
xk
e   x n1
x
e 
 Rn ( x), Rn ( x) 
,
(n  1)!
k  0 n!
where î is between 0 and x. If x   a, a , a  0 , then
e a  a n 1
Rn ( x ) 
 0, n   .
(n  1)!
This leads to the conclusion that the exponential function can be
written as a sum of a series:

xn
e x   , x   a, a .
n  0 n!
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
e
3
.

e  Tn (1)  Rn (1) 
( n  1)! ( n  1)!
It is sufficient to solve the inequality:
3
1

 (n  1)! 3000 .
(n  1)! 1000
This relation is fulfilled for n = 6. In conclusion
1 1 1 1 1
e  2       2.718,
2! 3! 4! 5! 6!

100

with an accuracy of 0.001.
The function f ( x )  sin x has derivatives of all order and

f ( k ) ( x )  sin( x  k ), k  IN; it follows that there exists   0, x  such
2
that
n

1
k k
x n 1
sin x   sin
x 
sin(  ( n  1) ) 
2
(n  1)!
2
k 0 k !


x x3 x5
xn
n
x n 1
 

 ....... 
sin

sin(  ( n  1) );
1! 3! 5!
n!
2 ( n  1)!
2
Here
m 1

T2 m  T2 m 1   ( 1) k
k 0

x 2 k 1
, m  IN ,
( 2k  1)!

and so
sin x  x 

x3 x5
x 2 m 1

 ....  ( 1) 2 m 1
 R2 m ( x ),
3! 5!
( 2m  1)!

where
R2 m ( x ) 


x 2 m 1
sin(x  ( 2m  1) ),   0,1, x  IR.
(2m  1)!
2

The function f ( x )  cos x expanded after Mac Laurin’s
formula has, in a similar way, the form:
x2 x4
n x n
x n 1
n
cos x  1 

 ....  cos


cos(x 
),  0,1 .
2! 4!
2 n! (n  1)!
2
The function f ( x )  shx has the form:
shx  x 

x3
x 2 m 1
f ( 2 m ) ( ) 2 m
 ....

x ,   0, x , m  IN, x  IR .
3!
(2m  1)!
(2m)!

The function f ( x )  chx , has the Mac Laurin`s polynomial:
T2 m ( x)  1 

x2
x 2m
 ....
 T2 m 1 ,
2!
(2m)!

and

101

sh
ch
x 2 m 1 , or R2 m 1 ( x ) 
x 2 m2 ,
( 2m  1)!
(2m  2)!
where m  IN ,   0, x , x  IR.
R2 m ( x ) 

The function f(x) = ln(x+1) has the extension:
x2 x3
( 1) n 1 n
( 1) n  x n 1
f ( x )  ln(1  x )  x 

 ..... 
x 
,
n
2
3
( n  1)(1  n ) n 1
x   1,  , n  IN ,   0,1.
The function f ( x)  (1  x) m has the extension
m m( m  1) 2
m(m  1)....( m  n  1) n
f ( x )  (1  x ) m  1  
x  ...... 
x 
1!
2!
n!
x ( x  1)....( x  n ) n 1

x (1  x ) m n 1 , x   1,  , m  IR,   0,1.
(n  1)!

Finding Limits with Taylor’s Formula
Example 6. Find l  lim
x 0





1
2 x 3  3 x 2  6 x  6 ln(1  x ) , by
x4

using Taylor’s formula.
Solution. Using the extension of logarithmic function we
know that there exists   0, x  , so that:

 6 4
x2 x3 x4
x5
x5
x
2 x 3  3x 2  6 x  6  x 
,





2
3
4 6(1  4 ) 6  4
(1  4 ) 6

therefore
 3
3
x
 .
l  lim  
6
x 0 2
(1  4 )  2


Example 7. Find l  lim
x 0

xshx  2chx  2
.
x 2 cos x  sin x 2

Solution. We will use the extension of elementary functions,
and so we will obtain the following:

102


 

x3
x2 x4
f ( x )  xshx  2chx  2  x  x 

 ....   2 
 ....   21 
3!
2! 4!

 

1 4

x  R4 ( x ),
2  3!
and


 

x2 x4
x6
 .....  

 ....    x 2 
g ( x )  x 2 cos x  sin x 2  x 2 1 
2! 4!
3!

 

4
x

 R4' ( x );
2!

according to Taylor’s formula there exists  ,  '  0, x  such that:
( ) 5
g ( 5) ( ' ) 5
x and R4' ( x) 
x .
5!
5!
In conclusion, the limit is:
R4 ( x ) 

f

(5)

1 4 f ( 5) ( ) 5
x 
x
1

2
3
!
5
!
 .
l  lim
( 5)
'
x 0
6
1
g ( ) 5
 x4 
x
2
5!

Exercise 8. Find





l  lim 3x  4 x 2  6 x 3  12 x 4  12 x 5 (ln(1  x )  ln x ) .
x

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

1
, and then we replace it in the given relation:
x

103


3 4
6 12 12
l  lim   2  3  4  5 ln(1  y ) 
y 0 y
y
y
y
y

y 0 
 lim
y 0

1
 lim 5
y 0 y






1
3 y 4  4 y 3  6 y 2  12 y  12 ln(1  y ) 
5
y
 4

y2 y3 y4 y5
y6
3
2
3 y  4 y  6 y  12 y  12( y  2  3  4  5  6(1  5 ) 5 ) 



12
, (  0,1).
5

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:
f ( x )  f ( a )  0, ( f ( x )  f ( a )  0), x  V  I .
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:
f ' ( x)  0 (*)
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
f : IR  IR, f ( x )  x 2 n 1 , n  IN , but x=0 is not an extreme point for f.

104

Theorem 2. Let f : I  IR be a function. Suppose that there
is a  I such that:
f ' (a )  f " (a )  ....  f ( n ) (a )  0 and f ( n 1) (a )  0, n  IN ,
and also f ( n 1) 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:
( x  a ) n 1 ( n 1)
f ( x)  f (a) 
f
( ) .
(1)
(n  1)!
Since f ( n 1) 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:
f ( n 1) ( x )  0, x  V  I , when f ( n 1) ( a )  0
(2)
or:
f ( n 1) ( x )  0, x V  I , when f ( n 1) ( a )  0
.
(3)
Case (a): n+1 is even:
(a1) If f ( n1) (a )  0 , then from (1) and (2) we have that:
f ( x )  f ( a )  0, x  V  I -{a},
since ( x  a) n 1  0 , therefore a is a local minimum point.
(a2) If f ( n1) (a)  0 , then from (1) and (3) we have that:
f ( x )  f ( a )  0, x  V  I ,
therefore a is a local maximum point.
Case (b): n+1 is odd:
Here ( x  a ) n1 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.
Exercise 9. Investigate the following functions, and
determine if they have, or not, local extreme points (maximum,
minimum).

105

f ( x )  sin x  x , g ( x)  sin x  x 

x3
,
3!

h( x)  4 x 2  3  2 cos 2 x .
Solution.1. For solving the equation (*): f ' ( x )  0 we have
f ' ( x )  cos x  1  0  xk  2k , k  ZI .
But
f " ( x k )   sin x k  0 and f ( 3) ( x k )   cos x k  1 .
It results that n+1 = 3 is not even; therefore all stationary points for f
are not local extremes.

2. We have:
x2
g ' ( x )  cos x  1 
 g " ( x )   sin x  x   f ( x )  0, x  IR .
2
It results that g' is a monotone increasing function, therefore a = 0 is
a unique stationary point. Moreover
g ' (0)  g " (0)  g (3) (0)  g ( 4) (0)  0 and g (5) (0)  1  0 .
In conclusion, n+1= 5 is an odd number and g has no local
extremes.
3. Since
h ' ( x )  8 x  4 sin 2 x  4 f ( 2 x ),
x=0 is the unique stationary point,
h ' (0)  h " (0)  h ( 3) (0)  0 and g ( 4) (0)  32  0 ,
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 Pn be 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
f  Pn , f ( x)  a 0  a1 x  ....  a n x n

106

is a vector written is the basis Bc, since f ( n 1) ( x )  0, x  IR , then:
f ' (a)
f ( n ) (a)
( x  a)  ..... 
( x  a) n
1!
n!
is the vector f written in the basis B.
f ( x)  Tn ( x)  f (a) 

Using Newton’s binomial formula:
( x  a )i  x i  Ci1 ax i 1  .....  ( 1) i 1 Cii 2 a i 2 x 2  ( 1) i 1 Cii 1a i 1 x 1  ( 1) i a i

for i  0,1,....n we can write the passing matrix:

TBc B

1 a
a2

1
 0 1  C2 a
0 0
C22

 ... ...
...

0
0 0
0 0
0



...
( 1) i a i
...
( 1) n a n

i 1 i 1 i 1
n 1 n 1 n 1
... ( 1) Ci a
... ( 1) Cn a 
... ( 1)i 2 Cii 2 a i  2 ... ( 1) n 2 Cnn 2 a n 2 


...
...
...
...

 Cn1 a n 1
...
0
...


...
0
...
1


Since for f ( x)  x i , we have
f ( k ) ( a )  i (i  1)(i  2).....(i  k  1) x i k  k! Cik x i k , k  i
and
f ( k ) (a)  0, k  i ;
then the expansion of f on powers of x-a (regarding Taylor’s
formula) is:
i

i

x i  Tn ( x )  Ti ( x )   ( x  a ) k   Cik a i  k ( x  a ) k 
k 0
2
i

k 0

 a  C ( x  a )  C ( x  a )  .....  Cii 1a ( x  a ) i 1  ( x  a ) i
i

1
i

2

and the inverse matrix is:

107

TBC1B  TBBc

 1
a
a2

1 C 21a
 0
 0
0
1

 ..... ..... .....

0
0
 0
 0
0
0


ai
an 
.....

..... Ci1a i 1 ..... C n1 a n 1 
..... Ci2 a i  2 ..... C n2 a n 2 
.
.....
.....
.....
...... 

.....
0
..... C nn 1a 
.....
0
.....
1 

.....

Exercise 10. Let f ( x)  x 3  2 x 2  5 x  1 and
B ={1,x+1,(x+1)2,(x+1)3} be a basis in P3/ IR. Write the polynomial
function f in the basis B.
Solution. We observe that, if we take a = -1, in Taylor’s
formula, since R3 ( x)  0 , we have:
3

f ( x)  T3 ( x)  
k 0

f

(1)
( x  1) k ;
k!

(k )

but:
f ( 1)  1  2  5  1  9, f ' ( 1)  3x 2  4 x  5 |x  1 
 3  4  5  12, f " ( x )  6 x  4 |x 1  10

and f ( 3) (1)  3 . In conclusion:
f ( x)  9  12( x  1)  5( x  1) 2  ( x  1) 3 .
Exercise 11. Find the inverse of the matrix:
 1  1 1  1


0 1  2 3 
T 
0 0
1  3


0 0
0
1 

using Taylor’s formula.
Solution. We notice that T  TBC B , where:



BC  1, x , x 2 , x 3



and B  1, x  1,( x  1 ) 2 ,( x  1 )3 .

Then: T 1  TBC1B  TBC B , hence:

108

T 1

1

0

0

0


1
1
0
0

1
2
1
0

1

3
;
3

1 

(since: x i  1  Ci1 ( x  1)  Ci2 ( x  1) 2  ...).

7. SOLVED PROBLEMS

Exercise 12. Expand the polynomial function:
f ( x)  x 5  x 5  2 x 2 ,
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:
5
f ( k ) (a)
f ( x)  T5 ( x)  
( x  a) k
!
k
k 0
where a  1 , and,in the second case a  1 . Then:
f (1)  0 ; f ( 1)  1  1  2  2 ;

f
f
f
f
f

'

( x)  5 x 4  4 x 3  4 x ,

''

( x )  20 x 3  12 x 2  4 ,

'''

( x )  60 x 2  24 x ,

(4)

( x )  120 x  24 ,

( 5)

( x )  5! ,

f
f

f
'''

'

(1)  5 ,

f

''

(1)  28 ;

(1)  84 ;

(4)

'

f

(1)  144 ;

f

'''

f

( 1)  5 ;

f

''

( 1)  12 ;

( 1)  36 ;
(4)

(1)  96 ;

hence
5

f ( x)  
k 0

f ( k ) (1)
4
( x  1) k  5( x  1)  14( x  1) 2  14( x  1) 3  6x  1 
k!

  x  1 ,
5

109

and
5

f ( x)  
k 0

f ( k ) ( 1)
( x  1) k  2  5( x  1)  6( x  1) 2  6( x  1) 3 
k!

 4( x  1) 4  ( x  1) 5 .
Exercise 13. Let f ( x)  x 2 e  x . Compute:
(a) f ( n ) ( x ) , for n  IN * .
(b) d 0 f , d 0 f ( x) , d 02 f , d 02 f ( x ) , d 03 f , d 03 f ( x ) .
(c) d12008 f , d12008 f (1) .
Solution. (a) Using Leibnitz’ formula, for n  2 , we have
that:
n

f ( n ) ( x )   Cnk ( e  x ) ( n k )  ( x 2 ) ( k ) 
e

x

(1)

k 0
n

x  ( 1)
2

n 1



n

C

k n 2

k
n

( 1) n k e  x ( x 2 ) ( k ) 



2nx  n (n  1)( 1) n 2 



= (1) n e  x x 2  2nx  n(n  1) ,
and we observe that this formula is true for n  IN .
(b) Generally d an f  f ( n ) ( a ) dx n and d an f ( x )  f n ( a )  x n , for
all n  IN .
Here:
d 0 f  f ' (o)dx  0 , d 0 f ( x)  0 ;
d 02 f  f ' ' (0) dx 2  2dx 2 , d 02 f ( x )  2 x 2 ;
d 03 f  f ' ' ' (0) dx 3  6dx 3 , d 03 f ( x )  6 x 3 .

(c) d12008 f  e 2008 (1  2  2008  2008  2007)dx 2008 ,
d12008 f (1)  4026041  e 2008 .
Exercise 14. Using Mac Laurin's formula compute:
12
l  lim 4 x 2  2  2chx
x 0 x
Solution. Since:





110

chx  1 

x2 x4 x5


sh(x) ,   (0,1) ,
2! 4! 5!

it follows that:

12  2
x2 x4 x5
x
2
2
(
1
sh(x)






4
x 0 x
2 24 5!



l  lim


12  x 4 2 x 5
 

sh(x)   1
4
x 0 x
5!
 12


 lim

Exercise 15. Find

7

130 with an accuracy of 10-6.
1
7

Solution. Let f : IR  IR, f ( x )  x , since 7 130  f (130) .
But 2 =128; therefore we shall use Taylor's formula with a=27=128
and we find n  IN such that:
f ( x)  Tn ( x)  Rn ( x)  10 6 ,
7

where Tn is Taylor's polynomial of n th degree and Rn the n th
remainder; we have x  130 , and, by Taylor’s theorem we know that
there exists an î  (128,130) such that:
Rn ( x )  Rn (130) 

f ( n 1) ( )
(130  128) ( n 1) .
( n  1)!

But:
1

1

2
1 1
1 1
f ( x)  x 7 ; f '' ( x )  (  1) x 7 ,
7
7 7
1
3
1 1
1
f ' ' ' ( x )  (  1)(  2) x 7 ,...,
7 7
7
6  7 ( k 1)

(k )
k 6  13    6  7( k  2) 
7
f ( x )  ( 1)
x
, for k  2 .
7k
'

Then:

2 n 1 6  13    6  7( n  1) 


Rn (130) 
( n  1)!
7 n 1
2 n 1 6  13    6  7( n  1) 1
1


 6 7 n  6
n 1
(n  1)!
7
2
10

111

6 7 n
7



We observe that n  2 verifies this inequality; therefore

2

7

130  T2 (130)  
k 0

2
 2  (128)
7

6

7

k

f k (128)
(130  128) 
k!
13


22 6
  2  128 7  2  0.0022322  0.000029  2.002203.
2! 7

Exercise 16. Prove that
1 1
1
1 1
1
1
2    .....   e  2    .....  
,
2! 3!
n!
2! 3!
n! n! n
and compute e with an accuracy of 10-5.
Solution. The exponential function can be written as a sum of
series:


n
xn
xn

 Rn ( x),
n  0 n!
k  0 k!
where, for x  1 , we have:
n
n
1
1

e

 Rn ( x ) .


k  0 k!
n  0 n!
But:

1
1
1


 ... 
Rn (1)  
( n  1)! ( n  2)!
m  n 1 m!

ex  




1
1
1 

 ... 
1

( n  1)!  n  2 ( n  2)( n  3)



1
1
1
1
1 
1



 .... 
1
2

1
( n  1)!  ( n  1) ( n  1)
n  n!
 ( n  1)! 1 
n 1
whence:
n
n
1
1
1
e 

n  n!
k  0 k!
k  0 k!
and the inequality is proved.
If we want to obtain the number e with an accuracy of 10-5 we
impose:


112

n

1
1
1

 5 ,
n  n! 10
k 0 k!
8
1
and we obtain for n  8 : e  T8 (1)    2.71828.
k 0 k!
0 e

Exercise 17. Write the vector
f ( x )  1  3x  3 x 2  x 3  IR 3 x 

in the base B  1, x  2, ( x  2) 2 , ( x  2) 3 .

Solution. Let a  2 in Taylor's formula. Then there exists
an  between 2 and x such that the remainder would be:
f ( 4 ) ( )
R4 ( x ) 
( x  2) 4  0 ,
4!
for all x  IR ; hence:
3

f ( x)  
k 0

k

f ( k ) (2)
( x  2) .
k!

But:
f ( 2 )  3; f ' ( x )  3  6 x  3 x 2 , f ' ( 2)  3;
f ' ' ( x )  6  6 x, f ' ' ( 2)  6, and
f ' ' '  3!
f  3  3( x  2)  3( x  2) 2  ( x  2) 3 .
In conclusion:
Exercise 18. Let f: I  IR→IR, be a function where I is an
interval and suppose that there exists f ( n1) (a )  IR where a  I.
Show that:
f ( x)  Tm ( x)
1

lim
f ( m 1) (a ), m  1,2,......, n

m
1
xa
(
m
1
)!

( x  a)
where Tm is Taylor's polynomial of degree m associated to f in the
point a.
Solution. Applying L'Hospital's rule, we have:

113

lim
x a

f ( x)  Tm ( x)
f ( m ) ( x)  f ( m ) (a)
1


lim
f ( m 1) (a ).
m 1
x  a ( m  1)  m...2( x  a )
(m  1)!
( x  a)

Exercise 19. Prove the inequality:
x 1 1

x x2

,
2
8

for all x>0.

Solution. Let f :  1,    R,
the Taylor's theorem :

f ( x) 

  0, x  so that f ( x )  f (0)  x 

x  1 . According to

f ' ' ( ) 2
x ,
2

hence:
3
x x2

x  1  1   1    2 .
2 8

Therefore
x 1 1

x x2

,
2
8

for all x