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Textbook by: Lawrence Baggett. E-mail the author

# Taylor Polynomials and Taylor's Remainder Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: This module contains taylor's remainder theorem, some remarks about Taylor series, and a test for local maxima and minima.

Definition 1:

Let ff be in Cn(Br(c))Cn(Br(c)) for cc a fixed complex number, r>0,r>0, and nn a positive integer. Define the Taylor polynomial of degree nn for ff at cc to be the polynomial TnT(f,c)nTnT(f,c)n given by the formula:

( T ( f , c ) n ) ( z ) = j = 0 n a j ( z - c ) j , ( T ( f , c ) n ) ( z ) = j = 0 n a j ( z - c ) j ,
(1)

where aj=f(j)(c)/j!.aj=f(j)(c)/j!.

REMARK If ff is expandable in a Taylor series on Br(c),Br(c), then the Taylor polynomial for ff of degree nn is nothing but the nnth partial sum of the Taylor series for ff on Br(c).Br(c). However, any function that is nn times differentiable at a point cc has a Taylor polynomial of order n.n. Functions that are infinitely differentiable have Taylor polynomials of all orders, and we might suspect that these polynomials are some kind of good approximation to the function itself.

## Exercise 1

Prove that ff is expandable in a Taylor series function around a point cc (with radius of convergence r>0r>0) if and only if the sequence {T(f,c)n}{T(f,c)n} of Taylor polynomials converges pointwise to f;f; i.e.,

f ( z ) = lim ( T ( f , c ) n ) ( z ) f ( z ) = lim ( T ( f , c ) n ) ( z )
(2)

for all zz in Br(c).Br(c).

## Exercise 2

Let fCn(Br(c)).fCn(Br(c)). Prove that f'Cn-1(Br(c)).f'Cn-1(Br(c)). Prove also that (T(f,c)n)'=T(f',c)n-1.(T(f,c)n)'=T(f',c)n-1.

The next theorem is, in many ways, the fundamental theorem of numerical analysis. It clearly has to do with approximating a general function by polynomials. It is a generalization of the Mean Value Theorem, and as in that case this theorem holds only for real-valued functions of a real variable.

## Theorem 1: Taylor's Remainder Theorem

Let ff be a real-valued function on an interval (c-r,c+r),(c-r,c+r), and assume that fCn((c-r,c+r)),fCn((c-r,c+r)), and that f(n)f(n) is differentiable on (c-r,c+r).(c-r,c+r). Then, for each xx in (c-r,c+r)(c-r,c+r) there exists a yy between cc and xx such that

f ( x ) - ( T ( f , c ) n ) ( x ) = f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 . ( 4 . 7 ) f ( x ) - ( T ( f , c ) n ) ( x ) = f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 . ( 4 . 7 )
(3)

REMARK If we write f(x)=Tf,cn)(x)+Rn+1(x),f(x)=Tf,cn)(x)+Rn+1(x), where Rn+1(x)Rn+1(x) is the error or remainder term, then this theorem gives a formula, and hence an estimate, for that remainder term. This is the evident connection with Numerical Analysis.

### Proof

We prove this theorem by induction on n.n. For n=0,n=0, this is precisely the Mean Value Theorem. Thus,

f ( x ) - T f , c 0 ( x ) = f ( x ) - f ( c ) = f ' ( y ) ( x - c . f ( x ) - T f , c 0 ( x ) = f ( x ) - f ( c ) = f ' ( y ) ( x - c .
(4)

Now, assuming the theorem is true for all functions in Cn-1((c-r,c+r)),Cn-1((c-r,c+r)), let us show it is true for the given function fCn((c-r,c+r)).fCn((c-r,c+r)). Set g(x)=f(x)-(T(f,c)n)(x)g(x)=f(x)-(T(f,c)n)(x) and let h(x)=(x-c)n+1.h(x)=(x-c)n+1. Observe that both g(c)=0g(c)=0 and h(c)=0.h(c)=0. Also, if xc,xc, then h(x)0.h(x)0. So, by the Cauchy Mean Value Theorem, we have that

g ( x ) h ( x ) = g ( x ) - g ( c ) h ( x ) - h ( c ) = g ' ( w ) h ' ( w ) g ( x ) h ( x ) = g ( x ) - g ( c ) h ( x ) - h ( c ) = g ' ( w ) h ' ( w )
(5)

for some ww between cc and x.x. Now

g ' ( w ) = f ' ( w ) - ( t ( f , c ) n ) ' ( w ) = f ' ( w ) - ( T ( f ' , c ) n - 1 ) ( w ) g ' ( w ) = f ' ( w ) - ( t ( f , c ) n ) ' ( w ) = f ' ( w ) - ( T ( f ' , c ) n - 1 ) ( w )
(6)

(See the preceding exercise.), and h'(w)=(n+1)(w-c)n.h'(w)=(n+1)(w-c)n. Therefore,

f ( x ) - ( T ( f , c ) n ) ( x ) ( x - c ) n + 1 = g ( x ) h ( x ) = g ' ( w ) h ' ( w ) = f ' ( w ) - ( T ( f ' , c ) n - 1 ) ( w ) ( n + 1 ) ( w - c ) n . f ( x ) - ( T ( f , c ) n ) ( x ) ( x - c ) n + 1 = g ( x ) h ( x ) = g ' ( w ) h ' ( w ) = f ' ( w ) - ( T ( f ' , c ) n - 1 ) ( w ) ( n + 1 ) ( w - c ) n .
(7)

We apply the inductive hypotheses to the function f'f' (which is in Cn-1((c-r,c+r)))Cn-1((c-r,c+r))) and obtain

f ( x ) - ( T ( f , c ) n ) ( x ) ( x - c ) n + 1 = f ' ( w ) - ( T ( f ' , c ) n - 1 ) ( w ) ( n + 1 ) ( w - c ) n = f ' ( n ) ( y ) n ! ( w - c ) n ( n + 1 ) ( w - c ) n = f ' ( n ) ( y ) ( n + 1 ) ! = f ( n + 1 ) ( y ) ( n + 1 ) ! f ( x ) - ( T ( f , c ) n ) ( x ) ( x - c ) n + 1 = f ' ( w ) - ( T ( f ' , c ) n - 1 ) ( w ) ( n + 1 ) ( w - c ) n = f ' ( n ) ( y ) n ! ( w - c ) n ( n + 1 ) ( w - c ) n = f ' ( n ) ( y ) ( n + 1 ) ! = f ( n + 1 ) ( y ) ( n + 1 ) !
(8)

for some yy between cc and w.w. But this implies that

f ( x ) - ( T ( f , c ) n ) ( x ) = f ( n + 1 ) ( y ) ( x - c ) n + 1 ( n + 1 ) ! , f ( x ) - ( T ( f , c ) n ) ( x ) = f ( n + 1 ) ( y ) ( x - c ) n + 1 ( n + 1 ) ! ,
(9)

for some yy between cc and x,x, which finishes the proof of the theorem.

## Exercise 3

Define f(x)=0f(x)=0 for x0x0 and f(x)=e-1/xf(x)=e-1/x for x>0.x>0. Verify that fC(R),fC(R), that f(n)(0)=0f(n)(0)=0 for all n,n, and yet ff is not expandable in a Taylor series around 0.0. Interpret Taylor's Remainder Theorem for this function. That is, describe the remainder Rn+1(x).Rn+1(x).

As a first application of Taylor's Remainder Theorem we give the following result, which should be familiar from calculus. It is the generalized version of what's ordinarily called the “second derivative test.”

## Theorem 2: Test for Local Maxima and Minima

Let ff be a real-valued function in Cn(c-r,c+r),Cn(c-r,c+r), suppose that the n+1n+1st derivative f(n+1)f(n+1) of ff exists everywhere on (c-r,c+r)(c-r,c+r) and is continuous at c,c, and suppose that f(k)(c)=0f(k)(c)=0 for all 1kn1kn and that f(n+1)(c)0.f(n+1)(c)0. Then:

1. If nn is even, ff attains neither a local maximum nor a local minimum at c.c. In this case, cc is called an inflection point.
2. If nn is odd and f(n+1)(c)<0,f(n+1)(c)<0, then ff attains a local maximum at c.c.
3. If nn is odd and f(n+1)(c)>0,f(n+1)(c)>0, then ff attains a local minimum at c.c.

### Proof

Since f(n+1)f(n+1) is continuous at c,c, there exists a δ>0δ>0 such that f(n+1)(y)f(n+1)(y) has the same sign as f(n+1)(c)f(n+1)(c) for all y(c-δ,c+δ).y(c-δ,c+δ). We have by Taylor's Theorem that if x(c-δ,c+δ)x(c-δ,c+δ) then there exists a yy between xx and cc such that

f ( x ) = ( T ( f , c ) n ) ( x ) + f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 , f ( x ) = ( T ( f , c ) n ) ( x ) + f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 ,
(10)

from which it follows that

f ( x ) - f ( c ) = k = 1 n f ( k ) ( c ) k ! ( x - c ) k + f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 = f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 . f ( x ) - f ( c ) = k = 1 n f ( k ) ( c ) k ! ( x - c ) k + f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 = f ( n + 1 ) ( y ) ( n + 1 ) ! ( x - c ) n + 1 .
(11)

Suppose nn is even. It follows then that if x<c,x<c, the sign of (x-c)n+1(x-c)n+1 is negative, so that the sign of f(x)-f(c)f(x)-f(c) is the opposite of the sign of f(n+1)(c).f(n+1)(c). On the other hand, if x>c,x>c, then (x-c)n+1>0,(x-c)n+1>0, so that the sign of f(x)-f(c)f(x)-f(c) is the same as the sign of f(n+1)(c).f(n+1)(c). So, f(x)>f(c)f(x)>f(c) for all nearby xx on one side of c,c, while f(x)<f(c)f(x)<f(c) for all nearby xx on the other side of c.c. Therefore, ff attains neither a local maximum nor a local minimum at c.c. This proves part (1).

Now, if nn is odd, the sign of f(x)-f(c)f(x)-f(c) is the same as the sign of f(n+1)(y),f(n+1)(y), which is the same as the sign of f(n+1)(c),f(n+1)(c), for all x(c-δ,c+δ).x(c-δ,c+δ). Hence, if f(n+1)(c)<0,f(n+1)(c)<0, then f(x)-f(c)<0f(x)-f(c)<0 for all x(c-δ,c+δ),x(c-δ,c+δ), showing that ff attains a local maximum at c.c. And, if f(n+1)(c)>0,f(n+1)(c)>0, then the sign of f(x)-f(c)f(x)-f(c) is positive for all x(c-δ,c+δ),x(c-δ,c+δ), showing that ff attains a local minimum at c.c. This proves parts (2) and (3).

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