

Applications
of differentiation  the graph of a function and its derivatives 
Taylor's
theorem (Taylor's
formula)  The extended mean value theorem 
The proof of Thaylor's theorem 
Maclaurin's
formula or Maclaurin's theorem 
Representing
polynomial using Maclaurin's and Taylor's formula 
Representing
polynomial using Maclaurin's and Taylor's formula examples 





Taylor's
theorem (Taylor's
formula)  The extended mean value theorem 
Suppose
f
is continuous
on the closed interval [x_{0},
x_{0}
+ h]
with continuous
derivatives to
(n

1)th
order on the
interval and its nth
derivative defined on
(x_{0},
x_{0}
+ h)
then, 

is called Taylor's theorem. 
To prove Taylor's theorem we
substitute x_{0}
= a,
x_{0}
+ h = b
and introduce the function 

where Q
is still undetermined and p
> 0. 
Since
j
(b) = 0
we
can define
Q
by setting j
(a)
= 0
too, 

then
we compute j
' (x)
to apply Rolle's
theorem. 
Note
that successive terms cancel (by the product rule) when differentiating j
(x), so we get 

Thus,
by Rolle's theorem there is some intermediate point c
inside the interval [a,
b]
such that j'
(c)
= 0, 

which
gives 


By
plugging the Q
back into above equation j
(a)
= 0
while returning substitutions, 
a
= x_{0},
b
= x_{0}
+ h,
b

a
= h
and where 



we
obtain Taylor's theorem to be proved. 
The
last term in Taylor's formula 

is
called the remainder and denoted R_{n}
since it follows after n
terms. 
By
plugging, a) p
= n into
R_{n}
we get 

the
Lagrange
form of the remainder, 
while if b) p
= 1
we get 

the
Cauchy
form of the remainder. 
Thus,
by substituting x_{0}
+ h = x
obtained is 

Taylor's
formula with Lagrange form of the remainder. 
Therefore, Taylor's
formula gives values of a function f
inside the interval [x_{0},
x_{0}
+ h]
using its value and the values
of its derivatives to
(n

1)th
order at the point x_{0} in the form 
f (x) =
P_{n 
}_{1}(x

x_{0}) + R _{n} 
where,
P_{n

}_{1}(x

x_{0})
is
(n

1)th
order Taylor polynomial for f
given by the first n
terms in the above
formula and R_{n}
is one of the given remainders. 

Maclaurin's
formula or Maclaurin's
theorem 
The
formula obtained from
Taylor's
formula by setting x_{0}
= 0


that
holds in an open neighborhood of the origin, is called Maclaurin's
formula or Maclaurin's
theorem. 

Consider
the polynomial
f_{n}_{
}(x)
= a_{n}x^{n}
+ a_{n}_{ }_{}_{
1}x^{n }^{}^{
}^{}^{
}^{1} + · · · + a_{3}x^{3}
+ a_{2}x^{2} + a_{1}x
+ a_{0}, 
let
evaluate the polynomial and its successive derivatives at the origin, 
f
(0) =
a_{0}, f '(0) = 1· a_{1},
f ''(0) = 1· 2a_{2},
f '''(0) = 1· 2· 3a_{3}, . . .
, f ^{(}^{n}^{)}(0) =
n!a_{n} 
we
get the coefficients, 


Therefore,
the Taylor polynomial of a function f
centered at x_{0}
is the polynomial of degree n
which has the same
derivatives as f
at x_{0},
up to order n. 
If
a
function f
is infinitely differentiable on an interval about a point x_{0}_{ }
or the origin, as are for example e^{x} and
sin
x,
then 
P_{0} (x)
= f (x_{0}), 
P_{1} (x) =
f (x_{0})
+ (x

x_{0})
f ' (x_{0}), 

P_{0},
P_{1},
P_{2},
. . . is a sequence of increasingly approximating polynomials for
f. 

Representing
polynomial using Maclaurin's and Taylor's formula 
If
given is an nth
degree polynomial 
P_{n}(x) =
a_{n}x^{n} + a_{n}_{ }_{}_{
1}x^{n }^{}
^{1} + ·
· · + a_{2}x^{2}
+ a_{1}x + a_{0 }then, 
P_{n}(0) =
a_{0},
P_{n}' (0) = a_{1},
P_{n}''(0) = 2!
a_{2},
P_{n}'''(0) = 3!
a_{3},
. . . , P_{n}^{(n)}(0) = n!
a_{n}
and
P_{n}^{(n
+ 1)} = 0 
so,
the coefficients of the polynomial 

therefore,
applying Maclaurin's formula, every polynomial can be written as 

since
P_{n}^{(n
+ 1)} = 0,
the remainder vanishes. 

Representing
polynomial using Maclaurin's and Taylor's formula, examples 
Example: Represent
the quintic y
= 2x^{5} + 3x^{4} 
5x^{3} + 8x^{2} 
9x + 1 using Maclaurin's
formula. 
Solution: Let write
all successive derivatives of the given quintic and evaluate them at the
origin, 
y' (x)
= 10x^{4} + 12x^{3} 
15x^{2} + 16x 
9,
y' (0)
= 
9 
y'' (x)
= 40x^{3} + 36x^{2} 
30x + 16,
y'' (0)
= 16 
y''' (x)
= 120x^{2} + 72x 
30,
y''' (0)
= 
30 
y^{IV}
(x)
= 240x + 72,
y^{IV}
(0)
= 72 
y^{V}
(x)
= 240,
y^{V}
(0)
= 240 
y^{VI}
= 0 and the last term of the polynomial a_{0}
= y(0)
= P_{5}(0) =
1, 
then
substitute obtained values into Maclaurin's
formula 


Example: Represent
the quartic y
= x^{4} 
4x^{3} + 4x^{2} + x 
4 at x_{0}
= 2 using
Taylor's
formula. 
Solution: Let write
all successive derivatives of the given quartic and evaluate them at x_{0}
= 2, 
y' (x)
= 4x^{3} 
12x^{2} + 8x + 1,
y' (2)
= 1 
y'' (x)
= 12x^{2} 
24x + 8,
y'' (2)
= 8 
y''' (x)
= 24x 
24,
y''' (2)
= 24 
y^{IV}
(x)
= 24,
y^{IV}
= 24 
y^{V}
(x)
= 0
and the last term, a_{0}
=
y(2)
= P_{4}(2) = 
2, 
then
substitute obtained values into Taylor's
formula 










Calculus contents
D 



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