03 truncation errors

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1 Part 3 Truncation Errors

Transcript of 03 truncation errors

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Part 3

Truncation Errors

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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

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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

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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

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A 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

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.

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• 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

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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.

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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

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Question

What will happen if we sum up only the first n+1 terms?

...)!1(!

...!3!2

1132

n

x

n

xxxxe

nnx

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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

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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

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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.

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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.

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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

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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

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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

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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

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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?

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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.

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Observation• A Taylor series converges rapidly near the

point of expansion and slowly (or not at all) at more remote points.

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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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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.

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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.

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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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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

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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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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

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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., 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.

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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

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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

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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

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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).

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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