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Find formulas for the Maclaurin polynomials p 0 , p 1 , p 2 and p 3 for f ( x ) = 1 1 + x . Find a formula for the n th Maclaurin polynomial. Write your anwer using sigma notation.

p 0 ( x ) = 1 ; p 1 ( x ) = 1 x ; p 2 ( x ) = 1 x + x 2 ; p 3 ( x ) = 1 x + x 2 x 3 ; p n ( x ) = 1 x + x 2 x 3 + + ( −1 ) n x n = k = 0 n ( −1 ) k x k

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Taylor’s theorem with remainder

Recall that the n th Taylor polynomial for a function f at a is the n th partial sum of the Taylor series for f at a . Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials { p n } converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to f . To answer this question, we define the remainder R n ( x ) as

R n ( x ) = f ( x ) p n ( x ) .

For the sequence of Taylor polynomials to converge to f , we need the remainder R n to converge to zero. To determine if R n converges to zero, we introduce Taylor’s theorem with remainder    . Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the n th Taylor polynomial approximates the function.

Here we look for a bound on | R n | . Consider the simplest case: n = 0 . Let p 0 be the 0th Taylor polynomial at a for a function f . The remainder R 0 satisfies

R 0 ( x ) = f ( x ) p 0 ( x ) = f ( x ) f ( a ) .

If f is differentiable on an interval I containing a and x , then by the Mean Value Theorem there exists a real number c between a and x such that f ( x ) f ( a ) = f ( c ) ( x a ) . Therefore,

R 0 ( x ) = f ( c ) ( x a ) .

Using the Mean Value Theorem in a similar argument, we can show that if f is n times differentiable on an interval I containing a and x , then the n th remainder R n satisfies

R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x a ) n + 1

for some real number c between a and x . It is important to note that the value c in the numerator above is not the center a , but rather an unknown value c between a and x . This formula allows us to get a bound on the remainder R n . If we happen to know that | f ( n + 1 ) ( x ) | is bounded by some real number M on this interval I , then

| R n ( x ) | M ( n + 1 ) ! | x a | n + 1

for all x in the interval I .

We now state Taylor’s theorem, which provides the formal relationship between a function f and its n th degree Taylor polynomial p n ( x ) . This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for f converges to f .

Taylor’s theorem with remainder

Let f be a function that can be differentiated n + 1 times on an interval I containing the real number a . Let p n be the n th Taylor polynomial of f at a and let

R n ( x ) = f ( x ) p n ( x )

be the n th remainder. Then for each x in the interval I , there exists a real number c between a and x such that

R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x a ) n + 1 .

If there exists a real number M such that | f ( n + 1 ) ( x ) | M for all x I , then

| R n ( x ) | M ( n + 1 ) ! | x a | n + 1

for all x in I .

Proof

Fix a point x I and introduce the function g such that

g ( t ) = f ( x ) f ( t ) f ( t ) ( x t ) f ( t ) 2 ! ( x t ) 2 f ( n ) ( t ) n ! ( x t ) n R n ( x ) ( x t ) n + 1 ( x a ) n + 1 .

We claim that g satisfies the criteria of Rolle’s theorem. Since g is a polynomial function (in t ), it is a differentiable function. Also, g is zero at t = a and t = x because

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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