Taylor's Theorem

Taylor Polynomial:  Pn(x) = f (x) + f '(x) (x - c) + f " (x) (x - c)2 + f '''(x) (x - c)3 + ...
                                                                               2!                       3!
+  f (n)(x)(x - c)n + ... , for n > 0
         n!

If c = 0, then Pn(x) is called a MacLaurin polynomial for f.
 

Taylor's Theorem with Remainder (also called the LaGrange form of the remainder)

If  f  has derivatives of all orders in an open interval I containing a, then for each x in I,

f(x) = f (a) + f '(a) (x - a) + f " (a) (x - a)2 + f '''(a) (x - a)3 + ... +  f (n)(a)(x - a)n + Rn(x) , n>0
                                               2!                       3!                                n!

    where     Rn(x) =   f (n+1)(c)   (x - a)n+1    for some c between a and x.
                                   (n + 1)!

Application:  One way to show that a series converges for all x is to show that Rn approaches 0 as n approaches 0  or

                     lim          (x - a)n+1   =  0
                     n®¥        (n + 1)!

PROOF:

To find Rn(x) we fix x in I (x ¹ a) and write  Rn(x) = f(x) - Pn(x)  where Pn(x) is the nth Taylor polynomial for f(x).  Then we let g be a function of t defined by

g(t) = f(x) = f(t) - f '(t)(x - t) - ... -   f (n)(t)  (x - t)n - Rn(x)  (x - t)n+1
                                                        n!                                (x - a)n+1

   (Note 1:  By the above definition of Rn(x), this last term is   f (n+1)(c) (x - a)n+1(x - t)n+1  =  f (n+1)(c)  (x - t)n+1
                                                                                                   (n + 1)!                 (x - a)n+1      (n + 1)!

The reason for defining g this way is that differentiation with respect to t has a telescoping effect.  For example, we have

  d     [ -f(t) - f '(t)(x - t)] = - f '(t) + f '(t) - f "(x - t) = - f "(x)(x - t) .  (Note 2:  x is a constant)
 dt

The result is that the derivative g’(t) simplifies to g'(t) = - f (n+1)(t) (x - t)n + (n + 1)Rn(x)(x - t)n
                                                                                           n!                                          (x - a)n+1
for all t between a and x.

Moreover, for a fixed x

g(a) = f(x) - [ Pn(x) + Rn(x) ] = f(x) - f(x) = 0
and
g(x) = f(x) - f(x) - 0 - ... - 0 = f(x) - f(x) = 0

Therefore, g satisfies the conditions of Rolle’s Theorem, and it follows that there is a number c between a and x such that
g’(c) = 0.  Substituting c for t in the equation for g’(t) and then solving for Rn(x), we obtain

 g'(c) = - f (n+1)(c) (x - c)n + (n + 1) Rn(x)  (x - c)n  = 0
                   n!                                             (x - a)n+1

           - f (n+1)(c) (x - c)n + (n + 1)  f (n+1)(c)   (x - a)n+1 (x - c)n    = 0  (Substituting in the value of Rn(x))
                   n!                                  (n + 1)!                      (x - a)n+1

           - f (n+1)(c) (x - c)n +   f (n+1)(c)    (x - c)n    = 0
                   n!                            n!
 

           so  Rn(x) =  f (n+1)(c) (x - a)n+1
                               (n + 1)!

Finally, since g(a) = 0, we have

 0 = f (x) - f (a) - f '(a)(x - a) - ... - f (n)(a) (x - a)n - Rn(x)
                                                        n!

f(x) = f(a) + f '(a)(x - a) + ... + f (n)(a) (x - a)n + Rn(x)
                                                   n!
Examples:

1.  Find the error involved in the MacLaurin polynomial P3 for y = sin x evaluated at x = 0.1 .

     P3 = x - x3/3!                                                                     (sin 0.1 = .0998334 ,  so  P3(0.1) < sin 0.1)
    sin 0.1 ~ P3(0.1) = .099833333333...

    sin x = x - x3/3! + R3(x)
    R3(x) = f (4)(c)  x4,  0 < c < 0.1
                  4!
   Since sin x is an increasing function on this interval,  sin 0 < sin c < sin 0.1

   0  <  sin c (0.1)4 <   1  (0.1)4  =  .000004166666...
              4!                  4!

   so    .99833333... - 0 < sin 0.1 < .09983333... + .000004166666...

          .99833333 < sin 0.1 < .0998375
 
 

2.  Determine n so Pn(x) < .001 for ln 1.2 expanded about c = 1 .

     f(x) = ln x;  f '(x) = 1/x ;  f "(x) = -1/x2 ; f '''(x) = 2/x3 ;  f (4)(x) = -6/x4 ;  f (5)(x) = 24/x5 ;  f (6)(x) = 120/x6 ;

     f (n) = (n - 1)!/xn

    Rn(x) =  f (n+1)(c) (1.2 - 1)n+1
                  (n + 1)!

    The maximum value of the (n + 1)st derivative of ln x is 1.  Substituting in the expression for the value of the nth derivative
     at c gives

     Rn(x) =        n!         (1.2 - 1)n+1  for  1 < c < 1.2
                 cn+1(n+1)!

     Rn(x) is largest when c is smallest, so let c = 1.

     Rn(x) =      1     (.2 )n+1   <    .001
                   (n+1)
                     (.2 )n+1   <    .001
                   (n+1)

                     1  n+1__   <      1
                 5n+1(n+1)         1000

                 1000 < 5 n+1(n + 1)  (Solve in the calculator)

                 n < 2.51159 , so n = 3

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