Post on 12-May-2015
1
Part 3
Truncation Errors
2
Key Concepts• Truncation errors
• Taylor's Series– To approximate functions – To estimate truncation errors
• Estimating truncation errors using other methods– Alternating Series, Geometry series,
Integration
3
Introduction
...),log(,,,),cos(),sin( xxxexx yx
on a computer using only +, -, x, ÷?
How do we calculate
One possible way is via summation of infinite series. e.g.,
...)!1(!
...!3!2
1132
n
x
n
xxxxe
nnx
4
Introduction
• How to derive the series for a given function?
• How many terms should we add?
or• How good is our approximation if we only sum
up the first N terms?
...)!1(!
...!3!2
1132
n
x
n
xxxxe
nnx
5
A general form of approximation is in terms of Taylor Series.
6
Taylor's Theorem: If the function f and its first n+1 derivatives are continuous on an interval containing a and x, then the value of the function at x is given by
nn
n
Raxn
af
axaf
axaf
axafafxf
)(!
)(
...)(!3
)()(
!2
)("))((')()(
)(
3)3(
2
x
a
nn
n dttfn
txR )(
!
)( )1(
where the remainder Rn is defined as
(the integral form)
Taylor's Theorem
7
1)1(
)()!1(
)(
nn
n axn
cfR
The remainder Rn can also be expressed as
Derivative or Lagrange Form of the remainder
for some c between a and x
(the Lagrange form)
The Lagrange form of the remainder makes analysis of truncation errors easier.
8
• Taylor series provides a mean to approximate any smooth function as a polynomial.
• Taylor series provides a mean to predict a function value at one point x in terms of the function and its derivatives at another point a.
• We call the series "Taylor series of f at a" or "Taylor series of f about a".
Taylor Series
nn
n
Raxn
af
axaf
axaf
axafafxf
)(!
)(
...)(!3
)()(
!2
)("))((')()(
)(
3)3(
2
9
Example – Taylor Series of ex at 0
...!
...!3!2
1
...)0(!
)0(...)0(
!2
)0(")0)(0(')0(
becomes 0at of seriesTaylor the,0With
.0any for 1)0( Thus
0any for )()(")(')(
32
)(2
)(
)(
n
xxxx
xn
fx
fxff
fa
kf
kexfexfexfexf
n
nn
k
xkxxx
Note:
Taylor series of a function f at 0 is also known as the Maclaurin series of f.
10
Exercise – Taylor Series of cos(x) at 0
0)0()sin()(1)0()cos()(
0)0()sin()(1)0(")cos()("
0)0(')sin()('1)0()cos()(
)5()5()4()4(
)3()3(
fxxffxxf
fxxffxxf
fxxffxxf
0
2
642
)(2
)!2()1(
...!6
0!4
0!2
01
...)0(!
)0(...)0(
!2
)0(")0)(0(')0(
becomes 0at of seriesTaylor the,0 With
n
nn
nn
n
x
xxx
xn
fx
fxff
fa
11
Question
What will happen if we sum up only the first n+1 terms?
...)!1(!
...!3!2
1132
n
x
n
xxxxe
nnx
12
Truncation Errors
Truncation errors are the errors that result from using an approximation in place of an exact mathematical procedure.
...)!1(!
...!3!2
1132
n
x
n
xxxxe
nnx
Approximation Truncation Errors
Exact mathematical formulation
13
How good is our approximation?
How big is the truncation error if we only sum up the first n+1 terms?
...)!1(!
...!3!2
1132
n
x
n
xxxxe
nnx
To answer the question, we can analyze the remainder term of the Taylor series expansion.
nn
n
Raxn
af
axaf
axaf
axafafxf
)(!
)(
...)(!3
)()(
!2
)("))((')()(
)(
3)3(
2
14
Analyzing the remainder term of the Taylor series expansion of
f(x)=ex at 0
1)1(
)()!1(
)(
nn
n axn
cfR
The remainder Rn in the Lagrange form is
for some c between a and x
For f(x) = ex and a = 0, we have f(n+1)(x) = ex. Thus
1
1
)!1(
]0[in somefor )!1(
nx
nc
n
xn
e
, xcxn
eR
We can estimate the largest possible truncation error through analyzing Rn.
15
ExampleEstimate the truncation error if we calculate e as
!7
1...
!3
1
!2
1
!1
11 e
This is the Maclaurin series of f(x)=ex with x = 1 and n = 7. Thus the bound of the truncation error is
481
177 106742.0
!8)1(
!8)!17(
eex
eR
x
The actual truncation error is about 0.2786 x 10-4.
16
ObservationFor the same problem, with n = 8, the bound of the truncation error is
58 107491.0
!9
eR
More terms used implies better approximation.
With n = 10, the bound of the truncation error is
710 106810.0
!11
eR
17
Example (Backward Analysis)
...!
...!3!2
132
n
xxxxe
nx
If we want to approximate e0.01 with an error less than 10-12, at least how many terms are needed?
This is the Maclaurin series expansion for ex
18
xnx exfexfcx )()(,01.00,01.0With )1(
To find the smallest n such that Rn < 10-12, we can find the smallest n that satisfies
121 10)01.0()!1(
1.1
n
n
Note:1.1100 is about 13781 > e
With the help of a computer:
n=0 Rn=1.100000e-02
n=1 Rn=5.500000e-05
n=2 Rn=1.833333e-07
n=3 Rn=4.583333e-10
n=4 Rn=9.166667e-13So we need at least 5 terms
1101.0
1 )01.0()!1(
1.1)01.0(
)!1()!1(
nnn
c
n nn
ex
n
eR
19
xnx exfexfcx )()(,5.00,5.0With )1(
Note:1.72 is 2.89 > e
With the help of a computer:
n=0 Rn=8.500000e-01
n=1 Rn=2.125000e-01
n=2 Rn=3.541667e-02
n=3 Rn=4.427083e-03
n=4 Rn=4.427083e-04
So we need at least 12 terms
n=5 Rn=3.689236e-05
n=6 Rn=2.635169e-06
n=7 Rn=1.646980e-07
n=8 Rn=9.149891e-09
n=9 Rn=4.574946e-10
n=10 Rn=2.079521e-11
n=11 Rn=8.664670e-13
Same problem with larger step size
115.0
1 )5.0()!1(
7.1)5.0(
)!1()!1(
nnn
c
n nn
ex
n
eR
20
To approximate e10.5 with an error less than 10-12, we will need at least 55 terms. (Not very efficient)
How can we speed up the calculation?
21
Exercise
If we want to approximate e10.5 with an error less than 10-12 using the Taylor series for f(x)=ex at 10, at least how many terms are needed?
xcxn
cfR
Rn
xxxe
Rxn
fx
fx
ffxf
xf
nn
n
n
n
nn
n
and 10 between somefor )10()!1(
)(
)!
)10(...
!2
)10()10(1(
)10(!
)10(...)10(
!2
)10(")10(
!1
)10(')10()(
is 10at )( of expansion seriesTaylor The
1)1(
210
)(2
The smallest n that satisfy Rn < 10-12 is n = 18. So we need at least 19 terms.
22
Observation• A Taylor series converges rapidly near the
point of expansion and slowly (or not at all) at more remote points.
23
Taylor Series Approximation Example:More terms used implies better approximation
f(x) = 0.1x4 - 0.15x3 - 0.5x2 - 0.25x + 1.2
24
Taylor Series Approximation Example:Smaller step size implies smaller error
f(x) = 0.1x4 - 0.15x3 - 0.5x2 - 0.25x + 1.2
Reduced step size
Errors
25
If we let h = x – a, we can rewrite the Taylor series and the remainder as
Taylor Series (Another Form)
nn
n
Rhn
afh
afhafafxf
!
)(...
!2
)(")(')()(
)(2
1)1(
)!1(
)(
n
n
n hn
cfR
h is called the step size.
h can be +ve or –ve.
When h is small, hn+1 is much smaller.
26
The Remainder of the Taylor Series Expansion
)()!1(
)( 11)1(
nnn
n hOhn
cfR
Summary
To reduce truncation errors, we can reduce h or/and increase n.
If we reduce h, the error will get smaller quicker (with less n).
This relationship has no implication on the magnitude of the errors because the constant term can be huge! It only give us an estimation on how much the truncation error would reduce when we reduce h or increase n.
27
Other methods for estimating truncation errors of a series
1. By Geometry Series
2. By Integration
3. Alternating Convergent Series Theorem
nn R
nnn
S
n ttttttttS ...... 3213210
Note: Some Taylor series expansions may exhibit certain characteristics which would allow us to use different methods to approximate the truncation errors.
28
Estimation of Truncation ErrorsBy Geometry Series
k
tkkkt
tktkt
tttR
n
n
nnn
nnnn
1
...)1(
...
...
1
321
12
11
321
k
tkR n
n
1
If |tj+1| ≤ k|tj| where 0 ≤ k < 1 for all j ≥ n, then
29
Example (Estimation of Truncation Errors by Geometry Series)
11.0
66
11
11
1
21
2
2
221
jt
t
jj
j
t
t
j
j
j
j
j
j
k
tkR n
n
1
jj
j
jt
jS2
2642 ......321
What is |R6| for the following series expansion?
66
66
10311.01
11.0
1
103,11.0
k
tkR
tk
n
Solution:
Is there a k (0 ≤ k < 1) s.t.|tj+1| ≤ k|tj| or |tj+1|/|tj| ≤ kfor all j ≤ n (n=6)?
If you can find this k, then
30
Estimation of Truncation ErrorsBy Integration
nn dxxfR )(
If we can find a function f(x) s.t. |tj| ≤ f(j) j ≥ n
and f(x) is a decreasing function x ≥ n, then
11321 )(...
njnjjnnnn jfttttR
31
Example (Estimation of Truncation Errors by Integration)
11
1133
jjj
13
1
)1(where
jttS jj
j
Estimate |Rn| for the following series expansion.
23 2
11
ndx
xR
nn
Solution:
We can pick f(x) = x–3 because it would provide a tight bound for |tj|. That is
So
32
Alternating Convergent Series Theorem (Leibnitz Theorem)
If an infinite series satisfies the conditions– It is strictly alternating.– Each term is smaller in magnitude than that
term before it.– The terms approach to 0 as a limit.
Then the series has a finite sum (i.e., converge) and moreover if we stop adding the terms after the nth term, the error thus produced is between 0 and the 1st non-zero neglected term not taken.
33
Alternating Convergent Series Theorem
16666.06
109.02ln
693.02ln
7833333340.05
1
4
1
3
1
2
11
,5 With
SR
S
n
Example 1:
Eerror estimated using the althernating convergent series theorem
Actual error
1
1432
)11()1(...432
)ln(1 of series Maclaurin
n
nn x
n
xxxxxS
x
34
Alternating Convergent Series TheoremExample 2:
77
8642
1076.2!10
11073.2)1cos(
5403023059.0 )1cos(
5403025793.0!8
1
!6
1
!4
1
!2
11
5, With
S
S
n
Eerror estimated using the althernating convergent series theoremActual error
0
12642
)!12()1(...
!6!4!21
)cos( of series Maclaurin
n
nn
n
xxxxS
x
35
ExerciseIf the sine series is to be used to compute sin(1) with an error less than 0.5x10-14, how many terms are needed?
...!17
1
!15
1
!13
1
!11
1
!9
1
!7
1
!5
1
!3
11)1sin(
171513119753
This series satisfies the conditions of the Alternating Convergent Series Theorem.
14102
1
)!32(
1
n
RnSolving
for the smallest n yield n = 7 (We need 8 terms)
Solution:R0 R1 R2 R3 R4 R5 R6 R7
36
Exercise
How many terms should be taken in order to compute π4/90 with an error of at most 0.5x10-8?
...4
1
3
1
2
11
90 444
4
405406)1(102
1
)1(3
1 83
nnn
Solution (by integration):
3
)1(
3
)1(
)1()1(
1...
33
4
1421
nx
dxxj
ttR
n
nnj
nnn
Note: If we use f(x) = x-3 (which is easier to analyze) instead of f(x) = (x+1)-3 to bound the error, we will get n >= 406 (just one more term).
37
Summary• Understand what truncation errors are
• Taylor's Series– Derive Taylor's series for a "smooth" function– Understand the characteristics of Taylor's Series
approximation– Estimate truncation errors using the remainder term
• Estimating truncation errors using other methods– Alternating Series, Geometry series, Integration