
Applications
of differentiation  the graph of a function and its derivatives 
Maclaurin's
formula or Maclaurin's theorem 
The approximation of the
exponential function by polynomial using Taylor's or Maclaurin's formula 
Properties
of the power series expansion of the exponential function 






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. 

The
approximation of the exponential function by polynomial using Taylor's
or Maclaurin's formula 
Example: Let
approximate the exponential function f
(x)
= e^{x} by
polynomial applying
Taylor's
or Maclaurin's
formula. 
Solution: The
exponential function is the infinitely differentiable function defined
for all real numbers whose 
derivatives
of all orders are equal e^{x}
so that, f (0)
= 1, f ^{(}^{n}^{)}(0)
= 1 and 

Where
Lagrange form 

of the
remainder 

so,
using Maclaurin's formula with Lagrange
remainder we get 


Since
for every real x, 

then 


therefore,
the function e^{x}
can be represented by the power series 

that
is absolutely convergent for all real x. 
The approximation of the exponential function by a sequence of polynomials is shown in the figure below. 


Properties
of the power series expansion of the exponential function 
Since
every polynomial function in the above sequence, f_{1
}(x),
f_{2 }(x),
f_{3}(x),
. . . , f_{n
}(x),
represents translation of its
original or source function, we calculate the coordinates of
translations, x_{0} and
y_{0},
to get their source forms. 
Let's
apply the method and formulas that are revealed and
explored under the 'Polynomial' sections. 
Thus, the coordinates of translations are, 

Note
that the above result proves the main property of the polynomial stating
that, an n^{th}
degree polynomial
function and all its successive derivatives to (n

1)^{th} order, have constant horizontal translation
x_{0}. 
Below
listed sequence of the polynomials and corresponding vertical
translations y_{0}
shows that their graphs
approach closer and closer to the point (x_{0},
y_{0}) or (1,
1/e) as n
tends to infinity. 

By
plugging the coordinates of translations with changed signs into the
polynomial expressed in the general 
form
y + y_{0}
= a_{n}(x + x_{0})^{n}
+ a_{n }_{}_{
1}(x + x_{0})^{n }^{}^{
1} + · · · + a_{2}(x +
x_{0})^{2} + a_{1}(x + x_{0})
+ a_{0}, 
after
expanding and reducing the expression, we get the source polynomial
function passing through the 
origin.
The above expression we can write as 
f_{s(n)}(x)
+ y_{0} = f_{n}(x + x_{0})
or f_{s(n)}(x)
= f_{n}(x + x_{0}) 
f_{n}(x_{0}), 
For
example, we obtain the source quadratic f_{s2}(x)
from the expression 
f_{s2}(x)
+ y_{0} = f_{2}(x + x_{0})
or f_{s2}(x)
= f_{2}(x + x_{0}) 
y_{0} that is, 

The
same way we get the source function of every polynomial as listed below. 

To
prove that expressions on the left and the right side of the same row
represent the same curve plug x_{0} and
y_{0}
into the source polynomial to get its translated form or, we can check
if the derivative
at x_{0}
=  1
of the
left side polynomial
is
equal to the derivative at x
= 0 of the source polynomial
(the graph of which passes
through the origin) on the right side. Thus,
for example 

Note
that all polynomials from f_{1}
to f_{n}
in the above sequence have the same horizontal translation x_{0}
=  1.

Recall
that an n^{th}
degree polynomial function and all its successive derivatives to (n

1)^{th} order, have constant horizontal translation
x_{0}. 
Since
every polynomial in the above sequence represents the derivative of its
successor, that is, 
f
'_{n}(x)
= f_{n }_{}_{
1}(x) and thus
f
'_{n}(x_{0})
= f_{n }_{}_{
1}(x_{0}). 
Therefore
as consequence, each xintercept
of odd polynomial in the sequence determines the abscissa of the
only extreme (minimum) of succeeding even polynomial and the abscissa of
the only point of inflection of succeeding
odd polynomial, as shows the picture above. 
In
the same way, for example the coefficients, a_{1},
a_{2},
and a_{3}
of the source polynomial f_{s5}(x)
are 

Hence,
the vertical translations y_{0}
of the successive derivatives are, 
f_{4}(x_{0})
= 1! f_{5}´(x_{0})
= 3/8, f_{3}(x_{0})
= 2! f_{5}´´(x_{0})
= 1/3 and
f_{2}(x_{0})
= 3! f_{5}´´´(x_{0})
= 1/2 , 
as
is shown above. 








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