Applications of differentiation - the graph of a function and its derivatives Taylor's theorem (Taylor's formula) - The extended mean value theorem
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
Taylor's theorem (Taylor's formula) - The extended mean value theorem
Suppose  f  is continuous on the closed interval [x0, x0 + h] with continuous derivatives to (n - 1)th order on the interval and its nth derivative defined on (x0, x0 + h) then, is called Taylor's theorem.
To prove Taylor's theorem we substitute  x0 = ax0 + 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 = x0,  b = x0 + 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 Rn since it follows after n terms.
By plugging,  a)   p = n  into Rn  we get the Lagrange form of the remainder,
while if     b)   p = 1    we get the Cauchy form of the remainder.
Thus, by substituting x0 + 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 [x0, x0 + h] using its value and the values of its derivatives to (n - 1)th order at the point x0 in the form
f (x) = Pn - 1(x - x0) + R n
where, Pn - 1(x - x0) is (n - 1)th order Taylor polynomial for f given by the first n terms in the above formula and Rn is one of the given remainders.
Maclaurin's formula or Maclaurin's theorem
The formula obtained from Taylor's formula by setting x0 = 0 that holds in an open neighborhood of the origin, is called Maclaurin's formula or Maclaurin's theorem.
Consider the polynomial   fn (x) = anxn + an - 1xn - - 1 + · · · a3x3 + a2x2 + a1x + a0,
let evaluate the polynomial and its successive derivatives at the origin,
f (0) = a0,     f '(0) = 1· a1,     f ''(0) = 1· 2a2,     f '''(0) = 1· 2· 3a3,  . . .  ,  f (n)(0) = n!an
 we get the coefficients, Therefore, the Taylor polynomial of a function f centered at x0 is the polynomial of degree n which has the same derivatives as f at x0, up to order n.
If a function f is infinitely differentiable on an interval about a point x0  or the origin, as are for example ex and  sin x, then
P0 (x) = f (x0),
P1 (x) = f (x0) + (x - x0) f ' (x0), P0, P1, P2, . . . 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) = ex  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 ex 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 ex 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,  f1(x)f2(x)f3(x), . . . ,  fn (x), represents translation of its original or source function, we calculate the coordinates of translations, x0 and  y0,  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 nth degree polynomial function and all its successive derivatives to (n - 1)th order, have constant horizontal translation x0.
Below listed sequence of the polynomials and corresponding vertical translations y0 shows that their graphs approach closer and closer to the point (x0, y0) 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 + y0 = an(x + x0)n + an - 1(x + x0)n - 1 + · · · + a2(x + x0)2 + a1(x + x0) + a0,
after expanding and reducing the expression, we get the source polynomial function passing through the origin. The above expression we can write as
fs(n) (x) + y0fn (x + x0)     or     fs(n) (x) =  fn (x + x0) - fn (x0),
For example, we obtain the source quadratic fs2(x) from the expression
fs2 (x) + y0f2 (x + x0)     or     fs2 (x) =  f2 (x + x0) - y0  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 x0 and y0 into the source polynomial to get its translated form or, we can check if the derivative at x0 = - 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 f1 to  fn in the above sequence have the same horizontal translation x0 = - 1.
Recall that an nth degree polynomial function and all its successive derivatives to (n - 1)th order, have constant horizontal translation x0.
Since every polynomial in the above sequence represents the derivative of its successor, that is,
f 'n(x) =  fn - 1(x)   and thus   f 'n(x0) =  fn - 1(x0).
Therefore as consequence, each x-intercept 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, a1, a2, and a3 of the source polynomial  fs5 (x) are Hence, the vertical translations y0 of the successive derivatives are,
f4(x0) = 1! f5´(x0) = 3/8,    f3(x0) = 2! f5´´(x0) = 1/3   and   f2(x0) = 3! f5´´´(x0) = 1/2 ,
as is shown above.   Calculus contents D 