LECTURE 1
Variations on Taylor's Formula
Numerical metho ds are ipso facto approximate metho ds. This b eing the case, it will b e imp ortant through-
out this course to determine the accuracy of numerical results. We shall b egin by reviewing the analytic
metho ds by whichwe approximate functions and howwe b ound the errors that arise from such approxima-
tions.
n
Definition 1.1. We denote by C [a; b] the set of functions on the interval [a; b] R that have continuous
n
derivatives up to order n. We denote by C R the set of functions on the real line that have continuous
1 1
derivatives up to order n. We denote by C [a; b] and C R the sets of functions for which derivatives of
al l orders exist on, respectively, [a; b] and R.
2 1 2
Example 1.2. If f x x sin 1=x then f is in C R but not in C R . To see this, note
1
2
=0 lim f x = lim x sin
x!0 x!0
x
1
2 0
since x ! 0 and sin is b ounded b etween 1 and 1. Thus f 2 C R .
x
df f x f 0
lim
x!0
dx x
x=0
1
x sin = lim
x!0
x
= 0
again b ecause it the limit of a function that is the pro duct of a function that vanishes at the limit p oint
1
and a function that remains b ounded. So f is also in C R . However,
2 0 0
d f f x f 0
lim
2
x!0
dx x
x=0
1
2x sin 1=x cos 0
x
= lim
x!0
x
1
cos
x
= lim
x!0
x
2
do es not exist. Thus, f=2 C R .
Remark 1.3. In general, wehave
1 3 2 1 0
C [a; b] C [a; b] C [a; b] C [a; b] C [a; b]
n+1
Theorem 1.4. Taylor's Theorem with Integral Remainder . If f 2 C [a; b], then for any points
x; x 2 [a; b],
o
n
X
1
k k
f x x x + R x f x =
o o n
k !
k =0
where
Z
x
1
n+1 n
R x f t x t dt
n
n!
x
o 1
1. VARIATIONS ON TAYLOR'S FORMULA 2
Proof. This theorem is surprisingly easy to prove. We start by using integration by parts to evaluate the
right hand side of the de nition of the error term R x : Setting
n
1
n n+1
u = x t dv = f t dt
n!
1
n 1 n
x t v = f t du =
n 1 !
and using the integration by parts formula
Z Z
x x
x
udv = uv j vdu
x
o
x x
o o
we nd
Z
x
x
1 1
n n n n 1
x t f t f t x t dt + R x
n
n! n 1 !
x
o x
o
1
n n
f x x x + R x =
o o n 1
n!
Wethus have a recursive formula for R x . Using this recursive formula over and over again we can reduce
n
the right hand side to
1 1
n n n 1 n 1
R x = f x x x f x x x +
n o o o o
n! n 1 !
1 1
0 n
f x x x + R x
o o 0
1! 0!
Z
n
x
X
1
n
0 0 n
f t x t dt = f x x x +
o o
0!
x
o
n=1
Z
n
x
X
n
0 n
= f t dt f x x x +
o o
x
o
n=1
n
X
n
n
= f x x x + f x f x
o o o
n=1
n
X
n
n
= f x x x + f x
o o
n=0
Solving the extreme sides for f x yields
n
X
1
k k
f x = f x x x + R x
o o n
k !
k =0
Remark 1.5. Note that in this formulation of Taylor's Theorem, wehave an explicit formula for computing
the error term R x . However, since the error term dep ends on the choice of x and x in a non-trivial way,
n o
we do not have at least at face value an understanding of how the error term changes as wevary x. Such
an understanding would b e critical if we are to regard
n
X
1
k k
f x f x x x
n o o
k !
k =0
as a p olynomial function approximate to the original function f x . Belowwe'll deduce b ounds on the
error term R x asx ranges throughout the interval [a; b].
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Theorem 1.6. Mean Value Theorem for Integrals . Let u and v becontinuous real-valued functions
on an interval [a; b], and suppose that v x 0 for al l x 2 [a; b]. Then there exists a point 2 [a; b] such
that
Z Z
b b
u x v x dx = u v x dx .
a a
1. VARIATIONS ON TAYLOR'S FORMULA 3
Proof. Let and denote, resp ectively, the minimal and maximal values of u x on[a; b]. Then
v x u x v x v x ; 8 x 2 [a; b]
Integrating this relationship b etween a and b yields
Z Z Z
b b b
v x dx u x v x dx v x dx
a a a
or
R
b
u x v x dx
a
R
b
v x dx
a
Nowbyhyp othesis, = u x and = u x for some x ;x 2 [a; b] . By the Intermediate Value Theorem
1 2 1 2
for Continuous Functions, for anynumber between and there must exist a p oint 2 [a; b] such that
u = . In particular, there must exist a p oint 2 [a; b] such that
R
b
u x v x dx
a
u =
R
b
v x dx
a
Hence there exists a p oint 2 [a; b] such that
Z Z
b b
v x dx u x v x dx = u
a a
Remark 1.7. The usual Mean Value Theorem; i.e., the statement that if f x iscontinuous and di eren-
tiable on [a; b] then there is a p oint 2 [a; b] such that
f b f a
0
f =
b a
0
is a sp ecial case of the formulation ab ove. To see this, note that if we take u x =f x and v x = 1, then
the ab ove theorem implies
Z Z
b b
0 0
f x dx = f dx
a a
Carrying out the integrations on b oth sides yields
f b f a
0 0
f b f a =f b a f =
b a
n n+1
Theorem 1.8. Taylor's Theorem with Lagrange Remainder . If f 2 C [a; b] and f x exists on
a; b then for any point x and x in [a; b]
o
n
X
1
k k
f x = f x x x + E x
o o n
k !
k =0
where
1
n+1
n+1
E x = f x x
n o
n + 1 !
for some point between x and x :
o
Proof. According to Taylor's Theorem with Integral Remainder is
Z
n
x
X
1 1
k k n+1 n
f x = f x x x + f t x t dt
o o
k ! n!
x
o
k =0
By the Mean Value Theorem for Integrals
Z Z
x x
1 1 1
n+1
n+1 n+1 n+1 n n
f f x x f t x t dt = x t dt =
o
n! n! n + 1 !
x x
o o
1. PROBLEMS 4
for some between x and x . Hence,
o
n
X
1 1
n+1
k k n+1
f x = f x x x + f x x
o o o
k ! n + 1 !
k =0
for some between x and x .
o
Remark 1.9. Note that in this formulation of Taylor's theorem the error term is still not precisely deter-
mined; b ecause the p oint in the theorem statement is left undetermined. We know only that there is some
p oint 2 [a; b] such that
n
X
1 1
n+1
k k n+1
f x x x = f x x f x
o o o
k ! n + 1 !
k =0
However, what we gain in this formulation are solid b ounds on the size of the error term over a range of x.
n+1
To see this let M b e the maximum value of f x on the interval [a; b], then
n
X
M 1
n+1
k k
f x x x jx x j f x
o o o
k ! n + 1 !
k =0
Note that this b ound is indep endent of the choice of x and x in [a; b].
o
n+1
Corollary 1.10. Alternative Form of Taylor's Theorem. Iff 2 C [a; b], then for any points x
and x + h 2 [a; b],
n
X
1
k k
f x + h = f x h + E h
n
k !
k =0
where
1
n+1 n+1
E h = f h
n
n + 1 !
in which the point lies between x and x + h.
1. Problems
1.1. Given that
n
d
n 1
ln jxj = 1 n 1 !
n
dx
x=1
a Use the Taylor Theorem with Integral Remainder to nd the magnitude of the error term R 1:99
100
when one approximates ln[1:99] using the rst 101 terms of the Taylor expansion ab out 1 of ln jxj.
b Use the Taylor Theorem with Lagrange Remainder to obtain an upp er b ound on the error term E x
100
when x ranges from 1.985 to 1.995 for the Taylor expansion of ln jxj ab out 1.