E. T. S. I. Caminos, Canales y Puertos1 Part 1 Engineering Computation.

E. T. S. I. Caminos, Canales y Puertos 1

Part 1

EngineeringComputation


Introduction

• Numerical methods are techniques which allow us to formulate mathematical problems to be solved by basic arithmetic operations.

• They are able to handle large systems of equations, non-linearities, complicated geometries and allow us solving engineering problems which have no analytical solution.

– The intelligent use of commercial software is favored by the knowledge of numerical methods.

– They allow us to design ad hoc programs to solve concrete problems.

– The allow us to become familiar with computers and understand the way they work.

– They are a good tool to reinforce the mathematical knowledge, because one of their aims consists in transforming complicate problems into simple arithmetic operations.


Introduction

• Roots of equations :– They are methods to solve

– They try to find the values of a variable for it to satisfy one equation.

– They are very useful in engineering projects, because in many occasions it is not possible to solve the design equations analytically.

.0)x(f\x 00


Introduction

• Systems of linear equations:

– They are methods looking for the set of values that simultaneously satisfy a system of algebraic equations.

– Calculus of structures,

electric circuits, supply networks, fit of curves, etc.

2222121

1212111

bxaxa

bxaxa


Introduction

• Optimization:

– Determine the value x0 leding to the optimal value of f(x).

– They can be subject to constraints.


Introduction

• Fitting curves. Fitting techniques can be divided into two groups:

– Regression. It is used when one has errors in the experimental data. One looks for the curve showing the trend of the data.

– Interpolation. It is used to fit tabulated data and predict intermediate values or extrapolated data.


Introduction

• Integration:

– Determine the area below a given curve.

– It has many applications in engineering. Calculus of centers of gravity. Calculus of areas, volumes, etc.

– It can also be used to solve differential equations.

b

a

dx)x(fI

f(x)

x

Integral

a b


Introduction

• Ordinary differential equations :

– They are very important because many problems can be stated in terms of variations and not in terms of magnitudes.

– There are two types of problems: Initial value problems, and boundary value problems.

dydt

= f (t;y)


Introduction

• Partial differential equations:

– Used for characterizing engineering problems where the behavior of the physical magnitude can be expressed in terms of speed change with respect to two or more variables.

– Approximation by finite differences or the finite element method.

)y,x(fy

u

x

u2

2

2

2

y

x


Mathematical Models

• A mathematical expression of a given model can be

• Analytic solution (t=0, v=0):

• Approximate solution:

DF

UF;maF )s/kgm(F 2 )kg(m )s/m(a 2

;mF

a ;mF

dtdv ;

mFF

dtdv UD

;m

cvmgdtdv )s/Kg(c

)tt()t(vmc

g)t(v)t(v i1iii1i


Mathematical Models

• To solve the problem numerically, one replaces the derivative by a divided finite difference, tus transforming the problem into a very simple one containing only simple algebraic operations:


Numerical Differentiation

Forward:

Centered:

How big a step size should we select? One- or two-sided formula:

What are the advantages of each? How is optimal step size affected by:

- precision of numerical calculations?- precision with which f is computed?- curvature of function f near x=1?- choice of formula?

i 1 ii

f x f xf x

h

i 1 i 1i

f x f xf x

2h


Numerical Methods Instead of solving for the exact solution we solve math problems with a series of arithmetic operations.

analytical solution: ln(b) – ln(a)numerical solution e. g., Trapezoidal Rule

Error Analysis (a) identify sources of error(b) estimate the magnitude of the error(c) determine how to minimize and control error

Example: dxb

a

1x

APPROXIMATION AND ERRORS


Mathematical Models

• Comparing solutions:

0 2 4 6 8 10 12 14 16 18 200

10

20

30

40

50

60

Numerical solution, t=1seg

Numerical solution, t=2seg

T (sec)

V (

m/s

ec)

Exact solution

Approximate Approximat(t=2s.) (t=1s.)

0 0 0 02 16,422 19,62 17,8193394 27,798 32,037357 29,6974396 35,678 39,896213 37,6151988 41,137 44,870026 42,893056

10 44,919 48,017917 46,41119512 47,539 50,010194 48,75633314 49,353 51,271092 50,31956616 50,611 52,069105 51,36159418 51,481 52,574162 52,05619320 52,085 52,893809 52,519203

Exactt(sec)


Approximations and Rounding Errors

• Unfortunately, computers introduce errors in calculations. However, since many engineering problems have not analytical solution, we are forced to used numerical methods (approximations). The only noption we have is to accept the error and try to reduce it up to a tolerable level.

• The only way of minimizing the errors is by knowing and understanding why they occur and how we can diminish them.

• The most frequent errors are:– Rounding errors, due to the fact that computers can work

only with a finite representation of numbers.– Truncation errors, due to differences between the exact and

the approximate (numeric) formulations of the mathematical problem being dealt with.

• Before centering in each one of them, we will see two important concepts on the computer representation of numbers.



• Significant figures of a number:– The significant figures of a number are those that can be

used with confidence.– The speedometer and the odometer in the figure estimate up

to three and seven significant figures, respectively, 49.5 and 87324.45.

– This concept has two important implications:

87324540

20

4060

80

100

120

6040

1. An approximation is acceptable when it is exact for a given number of significant figures.2. There are magnitudes or constants that cannot be represented exactly.

...123105.417

...14159265.3


Accuracy closeness of measured/computed values to the "true" value (vs. inaccuracy or bias)

Bias systematic deviation from truth, "general trend"

Precision closeness of measured/computed values with each other (spread or scatter), relates to the number of significant figures (vs. imprecision or uncertainty)



Approximations and Rounding Errors• Accuracy and precision:

– The errors associated with nd measurements can be characterized observing their accuracy and precision.

– Accuracy refers to how close the value is to the true value.

– Precision refers to how close are different measured values using the same method.

Numerical methods must be sufficiently exact (without bias) and precise to satisfy the requirements of engineering problems. From now on we will refer to error to refer to the inaccuracy and lack of precision of our predictions.


(a) inaccuratimprecise

(b) accurateimprecise

(c) Inaccurate precise

(d) Accurate precise




• Error definitions:– True value = approximation + absolute error.– Absolute error = true value - approximation .– Relative error = absolute error / true value .

– In real cases not always one can know the true value, thus:

– In many occasions, the error is calculated as the difference between the previous and the actual approximations.

%100valuetrue

errorabsolutet

%100valueeapproximat

erroraproximatea

%100ionapproximatactual

ionapproximatpreviousionapproximatactuala



– Thus, the stopping criterium of a numerical method can be:

– It is convenient to relate the errors with the number of significant figures.If the following relation holds, one can be sure that at least n significant figures are correct.

tolerancepercentprefixeds

sa

)%10*5.0( n2s



• Numerical systems:– A numerical system is a

convention to represent quantities. Since we have 10 fingers in our hands, our most familiis the numerical system which basis is 10. It uses 10 different digits.

– However, computers, due to the memory structure can only store two digits: 0 and 1. Thus, they use the binary system of numeric representation.

8 6 4 0 9

104 103 102 101 100

8 x 10000 = 800006 x 1000 = 60004 x 100 = 4000 x 10 = 09 x 1 = 9

86409

1 0 1 0 1

27 26 25 24 23

1 0 1

22 21 20

1 x 128 = 1280 x 64 = 641 x 32 = 320 x 16 = 161 x 8 = 81 x 4 = 40 x 2 = 01 x 1 = 1

173

a)

b)


Background: How are numbers stored in a computer?• The fundamental unit, a "word," consists of a string of

"bits" (binary digits).• Because computers are made up of gates or switches

which are either closed or open, we work in binary or base 2 system.

Example: An 8 bit word representation of the integer "35" is 00100011 or

0 0 1 0 0 0 1 1± 26 25 24 23 22 21 20

+ 0x26 1x25 0x24 0x23 0x22 1x21 1x20 =35

32 2 1Note: We can only represent a finite # of numbers; for our

case:–127 to +127 (127 = 27 – 1)or a total of 255 numbers (including 0)

Round-off Errors



• Representation of integer numbers:

– To represent base 10 numbers in binary form the signed magnitude method is used. The first digit stores the sign (0, positive and 1, negative). The remaining bits are used to store the number.

– A computer working with words of 16 bits can store integer numbers in the range -32768 to 32767.



• Floating point representation:– This representation is used for fractional quantities. It has

the fraction part, called mantissa, and an integer part, called exponent or characteristic.

– The mantissa is usually normalized, so that the value of m is limited:

eb*m

1mb1



• Hipothetical set:

– We assume a hipothetical set of floating point numbers for a machine using 7 bits as word.

– The smallest number that can be represented is 0.0625, and the largest is 7.



Sign. Sign.man. exp.

0 1 1 1 1 0 0 0,06250 1 1 1 1 0 1 0,07813 0,0156250 1 1 1 1 1 0 0,09375 0,0156250 1 1 1 1 1 1 0,10938 0,0156250 1 1 0 1 0 0 0,125 0,0156250 1 1 0 1 0 1 0,15625 0,031250 1 1 0 1 1 0 0,1875 0,031250 1 1 0 1 1 1 0,21875 0,031250 1 0 1 1 0 0 0,25 0,031250 1 0 1 1 0 1 0,3125 0,06250 1 0 1 1 1 0 0,375 0,06250 1 0 1 1 1 1 0,4375 0,06250 1 0 0 1 0 0 0,5 0,06250 1 0 0 1 0 1 0,625 0,1250 1 0 0 1 1 0 0,75 0,125

Exponent Mantissa Base 10 DifferenceSign. Sign.man. exp.

0 1 0 0 1 1 1 0,875 0,1250 0 0 1 1 0 0 1 0,1250 0 0 1 1 0 1 1,25 0,250 0 0 1 1 1 0 1,5 0,250 0 0 1 1 1 1 1,75 0,250 0 1 0 1 0 0 2 0,250 0 1 0 1 0 1 2,5 0,50 0 1 0 1 1 0 3 0,50 0 1 0 1 1 1 3,5 0,50 0 1 1 1 0 0 4 0,50 0 1 1 1 0 1 5 10 0 1 1 1 1 0 6 10 0 1 1 1 1 1 7 1

Base 10 DifferenceExponent Mantissa


EXAMPLE: “Our” Base 10 Machine:

This "Machine" carries three significant decimal digits and one exponent digit.

± 0.DDDE ± e

with: 0 < e < 9 100 < DDD < 999

We represent - 0.02778 in our machine:- 0.277E-1 w/ chopping t = 0.29%

- 0.278E-1 w/ rounding t = 0.07%

(most computers round, but some chop to save time)

Round-off Errors


Consequences of Computer Representation of #'s:

Limited Range of quantities can be represented because there are a finite # of digits in both the mantissa and the exponent.

smallest positive # 0.100E-9 or 10-10

largest positive # 0.999E+9 almost 109

Finite number of floating-point values can be represented within the above range.

± (0.100 to 0.999) E-9 to E+92 x 900 #'s x 19 #'s = 34,200 #'s

plus zero ( 0.000 E±e )

Thus, 34,201 #'s can be represented within the above range (zero appears only once).

Round-off Errors


The interval between successive represented numbers (quantization error) is not uniform, and increases as the numbers grow in magnitude.

- "quantization" error, Dx, is proportional to the magnitude x.

3.14x 100% 0.051%

0.312E 10.313E 1

- normalizing | x | by | x | gives an approximation of "machine epsilon"

- p, with both chopping and rounding, is represented

by 0.314E+1 with:"quantizing error" =

| x | = 0.01

0.312E 40.313E 4

| x | = 10.0

Consequences of Computer Representation of #'s



• Conclusions:– There is a limited range to represent quantities. The

limits correspond to “underflow” and “overflow”.

– There is a finite number of quantities that can be represented on a given range.



– The interval between numbers increases as the numbers grow in magnitude.

– This means that the errors are proportional to the magnitude of the number to be represented. Analyzing the relative error one gets:

– where is the machine epsilon, that is the worst possible relative error.

x

xwith chopping

2/x

x

With rounding


Machine Epsilon on "Our Base 10 Machine"

Try 0.01 - - - >0.100 E+1 + 0.100 E–1 = 0.101 E=1

1.0 + 0.01 = 1.01Try 0.00999 - - - >

0.100 E+1 + 0.999 E-2 = 0.100 E+11.0 + 0.00999 = 1.00999

after chopping ==> 1.00

Therefore, Machine epsilon is 0.01;

Quick formula: = b1- t = 101-3 = 0.01

t = # of sig. figs. in mantissa

Machine Epsilon: Smallest number such that: 1 + > 1

Machine Epsilon


Trouble happens when you add (or subtract) a small number to a large number:

For "our machine": 1,200 + 4 + 6 1,210

Working left-to-right: 1200 + 4 = 1204 which is chopped to 12001200 =0.120E+41200 + 6 = 1206which is chopped to 1200

Reverse the order:4 + 6 = 10 which is ok

10 + 1200 =1210 which is ok

Remedy: add or subtract small numbers together first.

Round-off Error due to Arithmetic Operations


Subtractive Cancellation (subtracting numbers of almost equal size) – too few significant figures left

Consider formulas such as :

Problems arise when we try to subtract two numbers of almost equal size

32,232 – 32,181 = 51 (Little precision left.)

Our Machine ("chopper"):0.322 E+5 – 0.321 E+5

32200 – 32100 = 100

With rounding:0.322 E+5 – 0.322 E+5

32200 – 32200 = 0

2b b 4acNeg. root

2a



Smearing Occurs when individual terms are larger than summation itself. Consider the exponential series with x = -10

Consider formulas such as:

With 7-decimal-digit accuracy:exact answer = 4.54 10-05

computed answer = – 6.26 10-05 (differs from

book value) (45 terms) wrong sign !

Largest intermediate terms are:9th = –2,755.732 & 10th = 2,755.732

2 3 4 5x x x x x

e 1 x2! 3! 4! 5!



Single Precision: 32 BIT WORDS 24 BITS assigned to mantissa (including the sign bit) 8 BITS to signed exponent

Double Precision: 64 BIT WORDS 56 BITS assigned to mantissa (including the sign bit) 8 BITS to signed exponent (not changed)

223 = 8,400,000 or almost 7 full decimal digits for single255 = 4x1016 or almost 17 decimal digits for double27 – 1 = 127; 2+127 ~ 2x1038 for both single and

double

Modern Computers (IEEE standard)


Error caused by the nature of the numerical technique employed to approximate the solution.

Example:

Maclaurin series expansion of ex

2 3 4 5x x x x x

e 1 x2! 3! 4! 5!

2x x

e 1 x2!

3 4 5x x x

3! 4! 5!

If we use a truncated version of the series:

Then the Truncation Error is:

Truncation Error



• Precautions:– Sums of large and small numbers: due to equaling the

exponent. They are common in sums of infinite series where the individual terms are very small when compared with the accumulated sum. This error can be reduced by summing first the small terms and using double precision.

– Cancellation of the subtraction: The subtraction of very similar numbers.

– Smearing: The individual terms are larger than the total sum.

– Inner products: They are prone to rounding errors. Thus, it is convenient to use double precision in this type of calculations.

n

1inn2211ii yxyxyxyx


Basic Idea:

Predict the value of a function, ƒ, at a point xi+1 based on the value of the function and all of its derivatives, ƒ, ƒ', ƒ",… at a neighboring point xi

Given xi, ƒ(xi), ƒ'(xi), ƒ"(xi), ... ƒn+1(xi),

we can predict or approximate ƒ(xi+1)

Taylor Series Expansion


General Form (Eq. 4.7 in C&C):

h = "step size" = xi+1 – xi

Rn = remainder to account for all other terms

= O (hn+1) with x not exactly known "on the order of hn+1 "

Note: f(x) must be a function with n+1 continuous derivatives

2 3 nn

i 1 i i i i i nh h h

f (x ) f (x ) hf (x ) f (x ) f (x ) f (x ) R2! 3! n!

n 1n 1h

f ( )(n 1)!

with xi xi+1



0th order T.S. approx. (n = 0): f(xi+1) = f(xi) + O (h1)

1st order T.S. approx. (n = 1): f(xi+1) = f(xi) + hf '(xi) + O (h2)

2nd order T.S. approx. (n = 2):

nth order T.S. approximation will be exact for an nth order polynomial

2 nn n 1

i 1 i i i ih h

f (x ) f (x ) h f (x ) f (x ) f (x ) (h )2! n!

O

2n 1

i 1 i i ih

f (x ) f (x ) h f (x ) f (x ) (h )2!

O



Zero orderFirst orderSecond order

f(xi )f(xi+1 ) f(xi )

f(xi+1 ) f(xi )+f '(xi )h

f(xi+1 ) f(xi )+f '(xi )h+ )+f "(xi )h2/2!

f(xi+1 )

True

f(x )

xi+1xi

h

x



Objective:Evaluate the derivatives of function, ƒ(xi), without doing it analytically.

When would we want to do this?1. function is too complicated to differentiate analytically:

2. function is not defined by an equation,

i.e., given a set of data points (xi, ƒ(xi)), i=1,…,n

i 0 1 2 3 4

xi 1.0 3.0 5.0 7.0 9.0ƒ(xi) 2.3 4.1 5.5 5.7 5.9

0.5x2 cos(1 x )e

1 0.5x

Numerical Differentiation from Taylor Series Expansion



– First derivative with backward difference.

)xx)(x('f)x(f)x(f 1iiii1i

)xx()x(f)x(f

)x('f1ii

1iii


Backward Difference Approx.:

First Derivative:

first backward difference

2

i 1 i i 1 i ih

f (x ) f (x ) (x x )f '(x ) f "( )2

2

i 1 i ih

f (x ) f (x ) hf '(x ) f "( )2

Letting h = xi - xi-1

2

i i i 1h

hf '(x ) f (x ) f (x ) f "( )2

i i 1f (x ) f (x )f '(x) (h)

h

O



Using data below calculate ƒ'(x1) :

i 0 1 2 3 4xi 1.0 3.0 5.0 7.0 9.0

ƒ(xi) 2.3 4.1 5.5 5.7 5.9

First Backward Finite-Divided-Difference at x1:

1 0f (x ) f (x )f '(x) (h)

h

O

14.1 2.3

f '(x ) (h)2

O

f ' (x1) 0.9 { + O (h) }

Example of 1st Backward FDD


Second Derivative:

2i 2 ii 2 i i 2 i i

x xf (x ) f (x ) x x f '(x) f "(x )

2!

+ O([xi-2– xi]3)

with h = xi– xi-1 and 2h = xi – xi-2

The 2nd order approximation to ƒ(xi-2) becomes:

ƒ(xi-2) = ƒ(xi) – 2hƒ'(xi) + 2h2 ƒ"(xi) +O (h3) [1]

2nd order approximation to ƒ(xi-1):2

3i 1 i i

h(x ) (x ) h '(x) "(x ) (h )

2!f f f f O [2]

Backward Difference Approximation


Subtracting 2*[2] from [1] yields:

f(xi-2) – 2f(xi-1) = –f(xi) + h2f"(xi) + O (h3)

Rearranging:

h2ƒ"(xi) = f(xi) – 2f(xi-1) + f(xi-2) + O (h3)

Second backward difference

3i i 1 i 2

i 2

f (x ) 2f (x ) f (x ) O(h )f "(x )

h

i i 1 i 2i 2

f (x ) 2f (x ) f (x )f "(x ) O(h)

h

Backward Difference Approximation


Using data below calculate ƒ"(x2) :i 0 1 2 3 4

xi 1.0 3.0 5.0 7.0 9.0ƒ(xi) 2.3 4.1 5.5 5.7 5.9

Second Backward Finite-Divided-Difference at x2:

2 1 02 2

f (x ) 2f (x ) f (x )f "(x ) (h)

h

O

2 2

5.5 2*4.1 2.3f "(x ) (h)

2

O

f " (5.0) - 0.1 { + O (h) }

Example of 2nd Backward FDD


What points are used for each form?

Backward:

…, ƒ(xi-2), ƒ(xi-1), ƒ(xi), ƒ(xi+1), ƒ(xi+2), …

Forward:


Centered:


Other Forms of Numerical Differentiation


Taylor Series and Truncation errors

- Higher order divided differences.

2iii1i

2iii1i

h!2

)x(''fh)x('f)x(f)x(f

h!2


- Second finite central divided difference


Forward:

i 1 ii

(x ) (x )f '(x ) (h)

h

O

i 2 i 1 ii 2

(x ) 2 (x ) (x )f "(x ) (h)

h

O

Centered:

2i 1 i 1i

(x ) (x )f '(x ) (h )

2h

O

2i 1 i i 1i 2

(x ) 2 (x ) (x )f "(x ) (h )

h

O

22 1i

- ( ) 4 ( ) -3 ( )f '(x ) = + O(h )

2hi i if x f x f x

Other Forms of Numerical Differentiation



• Use of the Taylor series to calculate derivatives.– First derivative with forward difference.

)xx)(x('f)x(f)x(f i1iii1i

)xx()x(f)x(f

)x('fi1i

i1ii



– First derivative with central differences.

h2)x(f)x(f

)x('f 1i1ii

2iii1i

2iii1i

h!2


h!2


3i

)3(

i1i1i h!3

)x(fh)x('f2)x(f)x(f


Questions:

• Which is a better approximation? Forward, Centered, or Backward?

• Why?

• When would you use which?

Note:

We also can get higher order forward, centered, and backward difference derivative approximations

[C&C Chapter 23, tabulated in Figs. 23.1-3]



Determine h to minimize the total error of a forward finite-divided difference approximation for:

i 1 if (x ) f (x )f '(x)

h

xi xi+1

• Round-off Error:

• Truncation Error:

i 1 if (x ) f (x ) hf '(x) f "( )

h 2

ˆx x x f f f with = machine epsilon.

. .i i(x h) (1 ) (x ) (1 ) hˆ "( )

h 2

f ' =

i i i(x h) (x ) 2 (x )RoundoffError h h

As a result:

Example Combining Roundoff and Truncation Error


Total error = truncation error + roundoff error

E = | Total Error | h

f "( )2

NOTE: Truncation error decreases as h decreases Round-off error increases as h decreases

i2 f (x )

h

+



To minimize total error E with respect to h, set the first derivative to zero:

i2

f "( ) 2 f (x )dE0

dh 2 h

Solve for h and approximate f "() as f

"(xi):

i

i

4 f (x )h

f "(x )



Using the first forward-divided-difference approximation with

error O(h) and a 5-decimal-digit machine:

e = b1-t = 101-5 = 10-4 = 0.0001

f '(x) = ; f "(x) = 0

i

i

4 f (x )h infinity

f "(x )

Linear Application:

Determine h that will minimize total error for calculating f’(x) for

f(x) = x at x = 1



f(x+h)= {exact: 3.1415} h (x+h) f(x+h)-f(x) [f(x+h)-f(x)]/h0 3.1415

0.000001 3.1415 0 00.00001 3.1416 0.0001 100.0001 3.1419 0.0004 4.00.001 3.1447 0.0032 3.20.01 3.1730 0.0315 3.150.1 3. 4557 0.3142 3.1421 6.2831 3.1416 3.146

Underlined digits are subject to round-off error. They are likely to be in error by ± one or two units. This does not cause much problem when h = 1, but causes large errors in the final result when h < 10-4.



Nonlinear Application:

Determine h for minimizing the total error for computing f’(x) for

ƒ(x) = ex at x = 3

Using the first forward-divided-difference approximation with

error O(h) and a 5-decimal-digit machine:

e = b1-t = 101-5 = 10-4 = 0.0001

f(x) = f '(x) = f "(x) = ex = 20.0855;

i

i

4 (x )h 0.02

"(x )

or about 0.01



Underlined digits subject to roundoff error.Bold digits in error due to truncation.

full precisionh f(x+h)=ex+h f(x+h)-f(x) [f(x+h)-f(x)] [f(x+h)-f(x)]

h h0 20.085 {exact = 20.085}

0.00001 20.085 0.0 0 20.0860.0001 20.087 0.002 20 20.0860.001 20.105 0.020 20 20.0960.01 20.287 0.202 20.2 20.180.1 22.198 2.113 21.13 21.121 54.598 34.513 34.523 34.512

Roundoff Truncation



Additional Error Terminology:

Error PropagationErrors which appear because we are basing current calculations on previous calculations which also incurred some form of error

Stability and Condition Number [see C&C 4.2.3]Numerically Unstable: Computations which are so sensitive to round-off errors that errors grow uncontrollably during calculations.Condition: sensitivity to such uncertainty; "well conditioned" vs. "ill conditioned"Condition Number: measure of the condition; i.e., extent to which uncertainty in x is amplified by ƒ(x)

C.N. 1 ===> "well-conditioned" C.N. >> 1 ===> "ill-conditioned"


Begin with Taylor Series:

where is an approximation of x.

Relative error of f(x) Relative error of x

f (x) f (x) f '(x)(x x)

f (x) f (x)

x x

x

The condition number is the ratio of the two relative errors:

f (x) f (x) f '(x)(x x)

x f '(x)Condition number (CN) =

f(x)

x

Condition Number


Condition Number

x f'(x)= f(x)

CN contrasts the uncertainty in x with the uncertainty in f(x)

C.N. 1 ===> "well-conditioned", i.e., The error in f(x) is similar to the error in x.

C.N. >> 1 ===> "ill-conditioned", i.e., the error in f(x) is amplified and small errors in x can produce large errors in f(x).

Condition Number


Condition Number

x f'(x)= f(x)

Example: Compute the condition number for f(x) = tan(x)

x f '(x)Condition number (CN) =

f(x)

f(x) = tan(x) = sin(x)/cos(x)

f'(x) = sec2(x) = 1/cos2(x)

CN = x/[cos(x) sin(x)]

This becomes large (ill-conditioned) when the denominator approaches zero, i.e., when x 0, /2, , …

Condition Number

E. T. S. I. Caminos, Canales y Puertos1 Part 1 Engineering Computation.

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Transcript of E. T. S. I. Caminos, Canales y Puertos1 Part 1 Engineering Computation.