Second Term 05/06 1
Part 3
Truncation Errors
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Key Concepts• Taylor's Series
• Using Taylor's series to approximate functions
• The Remainder of Taylor's Series
• Using the Remainder to estimate truncation errors
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Truncation Errors
Truncation Errors are those that result from using an approximation in place of an exact mathematical procedure.
ii
ii
tt
tvtv
t
v
dt
dv
1
1 )()(
...)!1(!
...!3!2
1132
n
x
n
xxxxe
nnx
Approximation of accelaration
MacLaurin series of ex
Approximation Truncation Errors
Exact mathematical formulation
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A more general form of approximation is in terms of Taylor Series.
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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
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1)1(
)()!1(
)(
nn
n axn
fR
The remainder Rn can also be expressed as
Derivative or Lagrange Form of the remainder
for some ξ between a and x
(the Lagrange form)
The Lagrange form of the remainder makes analysis of truncation errors easier.
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Any smooth function can be approximated 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.
Taylor Series
nn
iii
n
iii
iii
iiiii
Rxxn
xfxx
xf
xxxf
xxxfxfxf
)(!
)(...)(
!3
)(
)(!2
)("))((')()(
1
)(3
1
)3(
2111
Note:
This is the same expression except that a and x are replaced by xi and xi+1 respectively.
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If we let h = xi+1 - xi, we can rewrite the Taylor series and the remainder as
Taylor Series
nni
n
iiiii
Rhn
xf
hxf
hxf
hxfxfxf
!
)(...
!3
)(
!2
)(")(')()(
)(
3)3(
21
1)1(
)!1(
)(
n
n
n hn
fR
h is called the step size.
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Taylor Series ApproximationIf we take away the remainder, the Taylor series becomes the Taylor series approximation.
nii
in
iii
iii
iiiii
xxn
xfxx
xf
xxxf
xxxfxfxf
)(!
)(...)(
!3
)(
)(!2
)("))((')()(
1
)(3
1
)3(
2111
The remainder Rn accounts for all the terms not included in the approximation starting from term n+1.
n can be as small as zero.
The larger n is (more terms used), the better the approximation and thus the smaller the error.
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Zero-order approximation: use only the first term
)()( 1 ii xfxf
First-order approximation: use only the first two terms
))((')()( 11 iiiii xxxfxfxf
Second-order approximation: use only the first three terms
2111 )(
!2
)("))((')()( ii
iiiiii xx
xfxxxfxfxf
Some Terminologies
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Taylor Series Approximation Example:More terms used implies better approximation
f(x) = 0.1x4 - 0.15x3 - 0.5x2 - 0.25x + 1.2
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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
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The Remainder of the Taylor Series Expansion
1)1(
)!1(
)(
n
n
n hn
fR
Relationship between h and Rn
Smaller h implies smaller Rn, but if we reduce/magify h by a factor of k, how much smaller/larger Rn would be?
That is,
if h' = kh, what is Rn' in terms of Rn?
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Example: Relationship between h and R1
Use the first order Taylor series to approximate the function f(x) = x4 at x+h where x=1.
h True value
(x+h)4
1st Order Approx
f(x) + f'(x)h = x4+4x3h
= 1+4h
Error = R1=
True value –
1st order Approx.
1 16 5 11
0.5 5.0625 3 2.0625
0.25 2.441406 2 0.441406
0.125 1.601807 1.5 0.101807
0.0625 1.274429 1.25 0.024429
0.03125 1.130982 1.125 0.005982
0.015625 1.063980 1.0625 0.001480
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Pictorial View of h vs. R1
)log2(log 211 hRhR
1log.log Rvsh
A slop of 2 shown in the log-log plot of the remainder indicates that the truncation error is propotional to h2
)(62
12
12)(")(
!2
)("
22222
1
24
21
hOhhR
xxfxxf
hf
R
Theoritical result
Do they agree?
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The Remainder of the Taylor Series Expansion
)()!1(
)( 11)1(
nnn
n hOhn
fR
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.
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Example
...!
...!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
Note: Maclaurin series expansion is a special case of the Taylor series expansion with xi = 0 and h = x.
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1101.0
1 )01.0()!1(
1.1)01.0(
)!1()01.0(
)!1(
nnn
n nn
e
n
eR
xnx exfexfx )()(,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
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115.0
1 )5.0()!1(
7.1)5.0(
)!1()5.0(
)!1(
nnn
n nn
e
n
eR
xnx exfexfx )()(,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
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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.
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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
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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:
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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
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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
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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., convergent) 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.
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Alternating Convergent Series Theorem
16666.06
109.02log
693.02log
7833333340.05
1
4
1
3
1
2
112log
e
e
e
SR
S
Example:
77
8642
1076.2!10
11073.2)1cos(
5403023059.0)1cos(
5403025793.0!8
1
!6
1
!4
1
!2
11)1cos(
S
S
Eerror estimated using the althernating convergent series theoremAnother example:
Actual error
Actual error
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More ExamplesIf the sine series is to be used to compute sin(1) with an error less than 0.5x10-14, how many terms are needed?
...!9
1
!7
1
!5
1
!3
11)1sin(
9753
This series satisfies the conditions of the Alternating Convergent Series Theorem.
14102
1
!
1 n
Solving
for odd n, we get n = 17 (Need 8 terms)
Solution:
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More Examples
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
406102
1
3
1 83
nn
Solution (by integration):
3
44
1
4
3
1
3 n
xdxxk
nn
nk