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 [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 = x0b = x0 + hb - 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.
Representing polynomial using Maclaurin's and Taylor's formula
If given is an n-th degree polynomial
Pn(x) = anxn + an - 1xn - 1 + a2x2 + a1x + a0    then,
Pn(0) = a0Pn' (0) = a1Pn''(0) = 2! a2Pn'''(0) = 3! a3, . . . ,  Pn(n)(0) = n! an  and  Pn(n + 1) = 0  
so, the coefficients of the polynomial
therefore, applying Maclaurin's formula, every polynomial can be written as
since Pn(n + 1) = 0, the remainder vanishes.
Representing polynomial using Maclaurin's and Taylor's formula, examples
Example:  Represent the quintic  y = 2x5 + 3x4 - 5x3 + 8x2 - 9x + 1 using Maclaurin's formula.
Solution:  Let write all successive derivatives of the given quintic and evaluate them at the origin,
                    y' (x) = 10x4 + 12x3 - 15x2 + 16x - 9,           y' (0) = - 9
                    y'' (x) = 40x3 + 36x2 - 30x + 16,                    y'' (0) = 16
                    y''' (x) = 120x2 + 72x - 30                            y''' (0) = - 30
                    yIV (x) = 240x + 72                                       yIV (0) = 72
                    yV (x) = 240,                                                   yV (0) = 240
                    yVI = 0  and the last term of the polynomial  a0 y(0) = P5(0) = 1,  
then substitute obtained values into Maclaurin's formula
Example:  Represent the quartic  yx4 - 4x3 + 4x2 + x - 4 at x0 = 2 using Taylor's formula.
Solution:  Let write all successive derivatives of the given quartic and evaluate them at x0 = 2,
                    y' (x) = 4x3 - 12x2 + 8x + 1,              y' (2) = 1
                    y'' (x) = 12x2 - 24x + 8,                    y'' (2) = 8
                    y''' (x) = 24x - 24                            y''' (2) = 24
                    yIV (x) = 24                                           yIV = 24
                    yV (x) = 0     and the last term,   a0 y(2) = P4(2) = - 2,
then substitute obtained values into Taylor's formula
Calculus contents D
Copyright 2004 - 2020, Nabla Ltd.  All rights reserved.