Lecture 2 ENGR-516 Spring 2011

ENGR-516 Spring 2011

Adjunct Prof. Michael A. Soderstrand

[email protected]

405-334-8329

Lecture #2 January 24, 2011

1/24/2011 ENGR-516 Prof. Soderstrand Spring 2011 1

mailto:[email protected]


Homework #1 -- Due Monday January 24, 2011

Do the following problems from Chapters 1 and 2:

Problem 1.3, p. 21 of the text

Rather than the linear realationship of Eq. (1.7) you might choose to model the upward force on the parachutist as a second-order relationship.

Fu = -c'v2

where c' = a second-order drag coefficient (kg/m). a) Using calculus, obtain the closed-form solution for the case where the jumper is initially at rest (v=0 at t=0). b) Repeat the numerical calculation in Example 1.2 with the same initial condition and parameter values. Use a value of 0.225 kg/m for c'.




The velocity is equal to the rate of change of distance x(m),

dx/dt = v(t) [Eq P1.19]

a) Substitute Eq. (1.10) and develop and analytical solution for the distance as a function of time. Assume that x(0)=0. b) Use Euler's method to numerically integrate Eqs. (P1.19) and (1.9) in order to determine both the velocity and distance fallen as a function of time for the first 10s of free fall using the same parameters as in Example 1.2. c) Develop a plot of your numerical results together with analytical solutions.




Develop, debug, and document a program to determine the roots of a quadratic equation ax

2 + bx +

c, in either a high-level language or a macro language of your choice (MatLab strongly suggested). Use a subroutine procedure to compute the roots (either rel or complex). Perform test runs of the cases: a) a=1, b=6, c=2; b) a=0, b=-4, c=1.6; c) a=3, b=2.5, c=7.




The psuedocode below computes the factorial. Express this algorithm as a well-structured function in the language of your choice (MatLab is strongly recommended). Test it by computing 0! and 5! In addition, test the error trap by trying to evaluate -2!

Pseudocode

FUNCTION fac(n) IF n ≥ 0 THEN x=1 DOFOR i = 1,n x = x·i END DO fac = x ELSE display error message terminate ENDIF END fac


Chapter 3 – Approximation and Round-Off Errors

Errors are inherent in any numerical solution

Even when we have an exact analytic solution (as in the parachute or circuit example of Chapter 1),

as soon as we use a computer to calculate solutions, those solutions have error in them.

Often we do not have exact or analytic solutions, then numerical techniques give us approximations

– but how much error is there?


Sources of Error

This lecture will deal with all of the major errors associated with numerical analysis

Chapter 3 discusses accuracy and precision and errors due to number representation in the

computer.

Chapter 4 deals with truncation errors and in

detail errors associated with Taylor’s Series

approximations.


3.1 Significant Figures

Below is a car odometer (see text Fig 3.1)

How many significant digits are there?


There are eight significant digits.

However, only the first seven

can be used with confidence.

We can approximate the 8th

digit (126,462.25).


3.1 Significant Figures

Below is a car speedometer (see text Fig 3.1)

How many significant digits are there?


There are two significant digits.

However, only the first can be

used with confidence.

We can approximate the 2nd

digit (52).


3.2 Accuracy and Precision

Accuracy refers to how closely a

computed or measured value agrees with the true value.

Precision refers to how closely computed or measured values agree

with each other.


Illustration of Accuracy and Precision

Figure 3.2 p. 55 of text


Computer Precision

Precision on a computer comes from

the word-length used for computation.

Accuracy on a computer comes from

the quality of the algorithms used to

perform the calculations.


3.3 Error Definitions

Numerical errors include truncation

errors due to inexact mathematical operations and round-off errors due

to significant figure limitations.

True Value = Calculated Value +

True Error

Calculated Value is referred to in the

text as the Approximation


Error Definitions

Et = True Value – Approximation

Relative Error expressed as a percentage is given by:

𝜺𝒕 =𝑻𝒓𝒖𝒆 𝑬𝒓𝒓𝒐𝒓

𝑻𝒓𝒖𝒆 𝑽𝒂𝒍𝒖𝒆× 100% =

𝑬𝒕

𝑻𝑽× 100%

NOTE: This assumes we know the true

value.


Example 3.1

Problem Statement We measure a bridge and a rivet. We measure

10,000cm for the bridge and 11cm for the rivet. If the true values are

9,999cm and 10cm respectively, calculate a) the true error and b) the

relative percent error. (Note: This is

slightly different than in the text.)


Example 3.1

For the Bridge:

Et = 9999cm - 10000cm = -1cm

𝜺𝒕 =𝑬𝒕

𝑇𝑉=

−1𝑐𝑚

9999𝑐𝑚× 100% ≈ −.01%

For the Rivet:

Et = 10cm - 11cm = -1cm

𝜺𝒕 =𝑬𝒕

𝑇𝑉=

−1𝑐𝑚

10𝑐𝑚× 100% ≈ −10%


Approximate Error

Often we do not know the true value.

When we do not know the true value, we use the following equation to calculate the approximate error:

𝜺𝒂 =𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆 𝑬𝒓𝒓𝒐𝒓

𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏× 100%

In the above formula, the denominator is our

approximation and the numerator is the approximate error.

We know the approximation, but how do we find the approximate error?


Finding the Approximate Error

When we do not know the true value, it is often challenging to find the

approximate error.

However, in iterative algorithms, we can use the difference in successive

approximations as a reasonable indication of the approximate error:

𝜺𝒂 =𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏 − 𝑷𝒓𝒆𝒗𝒊𝒐𝒖𝒔 𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏

𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏× 100%

In the above formula, the denominator is our approximation and the numerator is the approximate error.


Sign of the Error

The error is negative if the approximation is larger than the

true value

The error is positive if the

approximation is smaller than the true value


Absolute Error

Often we are not concerned with the sign of the error.

In the rest of this class, we will use absolute error.

𝜺𝒂 = 𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆 𝑬𝒓𝒓𝒐𝒓

𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒊𝒐𝒏 × 100%


Error & Significant Digits

Often we want the error to be less than a certain number of significant

digits n.

The error will be less than n

significant digits if:

𝜺𝒂 < 𝜺𝒔 = 𝟎. 𝟓 × 𝟏𝟎𝟐−𝒏 %


Error & Significant Digits

Often we want the error to be less than a certain number of significant

digits n.

The error will be less than n

significant digits if:

𝜺𝒂 < 𝜺𝒔 = 𝟎. 𝟓 × 𝟏𝟎𝟐−𝒏 %

Let’s Take a

TEN MINUTE Break


Example 3.2 Error Estimates for Iterative Methods

Problem Statement. In mathematics, functions can often be represented by infinite series. For example, the exponential function can be computed using

𝑒𝑥 = 1 + 𝑥 +𝑥2

2+

𝑥3

3!+ ⋯ +

𝑥𝑛

𝑛!

Thus, as more terms are added to the sequence, the approximation becomes a better and better estimate of the true value of ex. This is called a Maclaurin series expansion.


Example 3.2 Error Estimates for Iterative Methods

Problem Statement Continued:

𝑒𝑥 = 1 + 𝑥 +𝑥2

2+

𝑥3

3!+ ⋯ +

𝑥𝑛

𝑛!

Starting with the simplest version, ex = 1, add terms one at a time to estimate e0.5 = 1.648721… Add terms until the absolute value of the approximate

error estimate a falls below a prespecified error

criterion s conforming to three significant figures.


Example 3.2 SOLUTION

First solve for the error-specification equivalent to three significant digits:

The true error can be calculated as:

𝜺𝒔 = 𝟎.𝟓 × 𝟏𝟎𝟐−𝒏 % = 𝟎.𝟓 × 𝟏𝟎𝟐−𝟑 % = 𝟎. 𝟎𝟓%

𝜺𝒕 =𝒆𝟎.𝟓 − 𝒂𝒑𝒑𝒓𝒐𝒙

𝒆𝟎.𝟓× 𝟏𝟎𝟎%


Example 3.2 SOLUTION

However, we usually do not know the true value and therefore cannot calculate the true error.

In such cases we must use the approximation error:

𝜺𝒂 =𝒂𝒑𝒑𝒓𝒐𝒙(𝒏) − 𝒂𝒑𝒑𝒓𝒐𝒙(𝒏 − 𝟏)

𝒂𝒑𝒑𝒓𝒐𝒙(𝒏)× 𝟏𝟎𝟎%


x = 0.5 s = 0.05%

e x = 1.64872127

n Approx t a

0 1.00000000 39.34693%

1 1.50000000 9.02040% 33.33333333%

2 1.62500000 1.43877% 7.69230769%

3 1.64583333 0.17516% 1.26582278%

4 1.64843750 0.01721% 0.15797788%

5 1.64869792 0.00142% 0.01579529%

6 1.64871962 0.00010% 0.00131626%

7 1.64872117 0.00001% 0.00009402%

8 1.64872127 0.00000% 0.00000588%


3.3.1 Iterative Calculations

Most of the methods in this course

are iterative using successive approximation to the true value

The computer implementation involves LOOPS usually ending when

the error drops below a specified value.


function [ v,ea,iter ] = IterMeth( x,es,maxit )

% Implements a general series expansion

% INPUT:

% (x,es,maxit): x = independent variable

% es = Specified error

% maxit = Maximum number of interations

% Function term(val,n) that defines the n-th term of the sequence

%

% OUTPUT:

% [v,ea,iter]: v = Approximate value for iter iterations

% ea = Approximate error after iter interations

% iter = number of iterations

iter=1; sol=1; ea=100;

while ea>es & iter < maxit,

solold=sol;

termn=term(x,iter);

sol=sol+termn;

iter=iter+1;

if sol ~= 0

ea=abs((sol-solold)/sol)*100;

end

end

v=sol;

return

end


function [ val ] = term( x,n )

% Function to calculate the

% n-th term of e^x

% INPUT: x, n

% OUTPUT: val

val=x^n/factorial(n);

return

end


3.4 Round-Off Errors

Round-off errors originate from the fact that computers retain only a fixed number of significant digits.

Irrational numbers cannot be expressed exactly.

Many numbers that are exact in decimal are not exact in binary used by computers.


3.4.1 Computer Representation of Numbers

Integers are represented in most computers in two’s compliment representation.

MatLab Integer Representations

Class Range of Values Name

Signed 8-bit integer -27 to 2

7-1 int8

Signed 16-bit integer -215

to 215

-1 int16


to 231

-1 int32


to 263

-1 int64

Unsigned 8-bit integer 0 to 28-1 uint8

Unsigned 16-bit integer 0 to 216

-1 uint16


-1 uint32


-1 uint64



Floating Point numbers are represented in most computers by IEEE Standard 754.

Name Common name Base Digits E min E max Decimal

digits

Decimal

E max

binary16 Half precision 2 10+1 -14 +15 3.31 4.51

binary32 Single precision 2 23+1 -126 +127 7.22 38.23

binary64 Double precision 2 52+1 -1022 +1023 15.95 307.95

binary128 Quadruple precision 2 112+1 -16382 +16383 34.02 4931.77

decimal32

10 7 -95 +96 7 96

decimal64

10 16 -383 +384 16 384

decimal128

10 34 -6143 +6144 34 6144



MatLab uses primarily double precision IEEE Standard 754 arithmetic.

Single precision is available, but half, quadruple, and the decimal representations are not.

Name Common name Base Digits E min E max Decimal

digits

Decimal

E max

binary16 Half precision 2 10+1 -14 +15 3.31 4.51

binary32 Single precision 2 23+1 -126 +127 7.22 38.23

binary64 Double precision 2 52+1 -1022 +1023 15.95 307.95

binary128

Quadruple

precision 2 112+1 -16382 +16383 34.02 4931.77

decimal32

10 7 -95 +96 7 96

decimal64

10 16 -383 +384 16 384

decimal128

10 34 -6143 +6144 34 6144


Binary and HEX

Binary HEX Binary HEX Binary HEX Binary HEX

0000 0 0100 4 1000 8 1100 C

0001 1 0101 5 1001 9 1101 D

0010 2 0110 6 1010 A 1110 E

0011 3 0111 7 1011 B 1111 F


3.4.1 Integer Representation in MatLab

uint8: 8-bit unsigned integer (0 – 255)

Binary HEX Decimal

0110 1010 6A 6161+10 = 106

0011 1011 3B 3161+11 = 59

1100 0101 C5 12161+5 = 197

1111 0111 F7 15161+7 = 247



uint8: 8-bit unsigned integer (0 – 255)

Decimal HEX Binary

16 1616 = 1R0 = 10 0001 0000

40 4016 = 2R8 = 28 0010 1000

228 22816 = 14R4 = E4 1110 0100

157 15716 = 9R13 = 9D 1001 1101



int8: 8-bit signed integer (-128 – 127)

Binary HEX Decimal

0110 1010 6A 6161+10 = 106

0011 1011 3B 3161+11 = 59

1100 0101 C5 4161+5-128 = -59

1111 0111 F7 7161+7-128 = -9



int8: 8-bit signed integer (-128 – 127)

Decimal HEX Binary

16 1616 = 1R0 = 10 0001 0000

40 4016 = 2R8 = 28 0010 1000

-28 (256-28)16 = 14R4 = E4 1110 0100

-99 (256-99)16 = 9R13 = 9D 1001 1101



int16: 16-bit signed integer (-32,768 – 32,767)

Binary HEX Decimal

0110 1010 0011 1101 hex2dec(‘6A3B’)

6A3B 6163+10162

+3161+11 = 27,195 1100 0101 1111 0111

hex2dec(‘45F7’)+ intmin(‘int16’)

C5F7 4163+5162

+15161+7-32768 = -14,857




Decimal HEX Binary

4136 413616 = 258R8 = 1028

25816=16R2

1616=1R0

0001 0000 0010 1000

-7011 (65,536-7011)16 = 3657R13 = E49D

365716 = 228R9

22816 = 14R4

1110 0100 1001 1101




Decimal HEX Binary

4136 413616 = 258R8 = 1028

25816=16R2

1616=1R0

0001 0000 0010 1000

-7011 (65,536-7011)16 = 3657R13 = E49D

365716 = 228R9

22816 = 14R4

1110 0100 1001 1101

Let’s Take a

TEN MINUTE Break


3.4.1 Floating Point Representation

Computer equivalent of scientific notation.

𝐼 = 2,356 𝐹 = 2.356 × 103 (normalization)

𝐼 = 2,356 𝐹 = 2.36 × 103 (norm & round)

Binary equivalent:

𝐼 = 1011 𝐹 = 1.011 × 23 (normalization)

𝐼 = 1011 𝐹 = 1.10 × 23 (norm & round)

In binary, we save a bit, because the whole number is always 1 – we only need to store the fractional part.


IEEE 754 Double Precision

MSB is the sign of the number (0 = +; 1 = -)

The exponent is 11 bits in a biased number system (subtract 1023 from the actual number)

The mantissa is formed by 1 + the 52-bit fraction. (53 bits total)


Exponent encoding

The double precision binary floating-point exponent is encoded using an offset binary representation, with the zero offset being 1023; also known as exponent bias in the IEEE 754 standard. Examples of such representations would be: ■ Emin (1) = -1022 ■ E (50) = -973 ■ Emax (2046) = 1023 Thus, as defined by the offset binary representation, in order to get the true exponent the exponent bias of 1023 has to be subtracted from the written exponent. The exponents 0x000 and 0x7ff have a special meaning: ■ 0x000 is used to represent zero (if F=0) and subnormals (if F≠0); and ■ 0x7ff is used to represent infinity (if F=0) and NaNs (if F≠0), where F is the fraction mantissa. All bit patterns are valid encoding.



(−𝟏)𝒔𝒊𝒈𝒏 × 𝟐(𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕−𝟏𝟎𝟐𝟑) × 𝟏. 𝒎𝒂𝒏𝒕𝒊𝒔𝒔𝒂

0x3ff0 0000 0000 0000 = 1

0x3ff0 0000 0000 0001 = 1.0000000000000002,

the next higher number > 1

0x3ff0 0000 0000 0002 = 1.0000000000000004

0x4000 0000 0000 0000 = 2

0xc000 0000 0000 0000 = –2




0x0000 0000 0000 0000 = 0

0x8000 0000 0000 0000 = –0

0x7ff0 0000 0000 0000 = Infinity

0xfff0 0000 0000 0000 = -Infinity




0x0000 0000 0000 0001 ≈ 4.9406564584124654 x 10-324

(Min subnormal positive double)

0x0010 0000 0000 0000 ≈ 2.2250738585072014 x 10-308

(Min normal positive double)

0x7fef ffff ffff ffff ≈ 1.7976931348623157 x 10308

(Max Double)

Note: Subnormals fill the gap between zero and smallest number (see text pp. 65-67 for explanation of this gap).




0x7ff0 0000 0000 0001 = qNaN

0xfff8 0000 0000 0000 = sNaN


There are three kinds of operations which return NaN

Operations with a NaN as at least one operand

Indeterminate forms o The divisions 0/0, ∞/∞, ∞/−∞, −∞/∞, and −∞/−∞ o The multiplications 0×∞ and 0×−∞ o The additions ∞ + (−∞), (−∞) + ∞ and equivalent subtractions o The standard has alternative functions for powers:

The standard pow function and the integer exponent pown function define 00, 1∞, and ∞0 as 1.

The powr function defines all three as invalid operations (NaN).

Real operations with complex results, for example: o The square root of a negative number o The logarithm of a negative number o The inverse sine or cosine of a number which is less than −1 or greater

than +1.


3.4.2 Arithmetic Manipulations

Floating-point addition and subtraction can cause significant error.

Exponents must be the same to add or subtract

1.557 104

+ 4.381 102

Can’t Add!

The mantissa of the number with the smaller exponent is modified to make the exponents the same

1.557 104

+ 0.04381 104

1.60081 104



1.557 104

+ 0.04381 104

1.60081 104

Exponents However, we must now either truncate (chop) or round to the machine precision (4 digits)

1.557 104

+ 0.04381 104

1.600 104 Chopped

1.557 104

+ 0.04381 104

1.601 104 Rounded



Floating-point subtraction of two numbers that are very close to each other causes major problems.

Exponents must be the same to add or subtract

7.642 103

- 7.641 103

0.001 103

Renormalization creates three non-significant digits

7.642 103

- 7.641 103

1.000 100


Example 3.7

Investigate the effect of round-off error on large numbers of interdependent computations.

The text suggests adding 1 and .00001 in single precision to themselves 10,000 times.

The first sum is 10,000 but the second will not be 1 due to round-off errors.

The text then suggest doing the second sum in double precision.


Example 3.7 Generalized % Program Fig0312 (Page 72 of text)

% INPUTS: x1 = the first number to add (eg: 1)

% x2 = the second and third number to add (eg: 0.1)

% x3 = the second and third number to add (eg: 0.1)

% n = numer of iterations (eg: 10,000,000)

% OUTPUTS: sum1 = x1 summed n times (single precision)

% sum2 = x2 summed n times (single precision)

% sum2 = x3 summed n times (double precision)

sum1=single(0);

sum2=sum1;

sum3=0;

for i = 1:n

sum1=sum1+x1;

sum2=sum2+x2;

sum3=sum3+x3;

end


Example 3.7 Generalized % Program Fig0312 (Page 72 of text)

sum1=single(0);

sum2=sum1;

sum3=0;

for i = 1:n

sum1=sum1+x1;

sum2=sum2+x2;

sum3=sum3+x3;

end

t1=sprintf('n = %0.10g, ',n);

t2=sprintf('x1 = %0.8g, x2 = %0.8g, x3 = %0.8g\n',x1,x2,x3);

t3=sprintf('sum1 = %0.8g\n',sum1);



t=sprintf('%s%s%s%s%s',t1,t2,t3,t4,t5);

sprintf(t)

n = 10000000, x1 = 1, x2 = 0.1, x3 = 0.1 sum1 = 10000000 sum2 = 1087937 sum3 = 1000000


Example 3.8

Find the roots of the quadratic equation with a=1, b=30000000.0000001, c=3. (x+30000000)(x+0.0000001)

MatLab and Excel have trouble with this

EDU>> a=1;b=30000000.0000001;c=3; EDU>> r1=(-b+sqrt(b^2-4*a*c))/2/a

r1 = -1.00582838058472e-007

EDU>> r2=(-b-sqrt(b^2-4*a*c))/2/a

r2 = -30000000

EDU>> r1=2*c/(-b-sqrt(b^2-4*a*c))

r1 = -1e-007

a = 1

b = 30000000.0000001

c = 3

r1 = -0.000000100582838

r2 = -30000000

r1 = 0.000000100000000


Example 3.9

Using a Taylor’s Series to calculate ex when x is negative is not recommended.

Instead calculate 1/e|x|.

However, the example in the text on pp. 74-75 works fine in both MatLab and Excel.

You need to use a higher value of x to see the effect (try -19 and -20).

EDU>> IterMeth(x,es,maxit) ans = 2.55376446514562e-009 EDU>> exp(-19) ans = 5.60279643753727e-009


Example 3.9

Using a Taylor’s Series to calculate ex when x is negative is not recommended.

Instead calculate 1/e|x|.

However, the example in the text on pp. 74-75 works fine in both MatLab and Excel.

You need to use a higher value of x to see the effect (try -19 and -20).

EDU>> IterMeth(x,es,maxit) ans = 2.55376446514562e-009 EDU>> exp(-19) ans = 5.60279643753727e-009

Let’s Take a

TEN MINUTE Break


Chapter 4 – Truncation errors and the Taylor Series

Truncation Errors are those that result from using

an approximation in place of an exact mathematical procedure

Euler’s Method is an example of a first order approximation to the real next value of a function.

Most approximation methods including Euler’s Method make use of a Taylor Series Approximation.


4.1 – The Taylor Series

If a function f and its first n+1 derivatives are continuous on an interval containing x and x+h:

𝑓(𝑥 + ℎ) = 𝑓(𝑥) + 𝑓 ′(𝑥)ℎ +𝑓′′ (𝑥)

2!ℎ2 + ⋯ +

𝑓 𝑛 (𝑥)

𝑛 !ℎ𝑛 + 𝑅𝑛

Where the nth-order approximation error is give b:.

𝑅𝑛 = (𝑥 − 𝑡)𝑛

𝑛!𝑓 𝑛+1 (𝑡)𝑑𝑡 =

𝑓 𝑛+1 ()

(𝑛 + 1)!ℎ𝑛+1

𝑥+ℎ

𝑡=𝑥

Where is some value between x and x+h


4.1 – Zero, 1st and 2nd Order Approximations

Rn is the slope to the exact prediction:


Example 4.1 – Exact Taylor Series Approximation

For polynomials of order n, the n-th order Taylor Series

approximation is exact

Consider 𝑓(𝑥) = −0.1𝑥4 − 0.15𝑥3 − 0.5𝑥2 − 0.25𝑥 + 1.2 (n=4)

𝑓 ′(𝑥) = −0.4𝑥3 − 0.45𝑥2 − 𝑥 − 0.25

𝑓 ′′ (𝑥) = −1.2𝑥2 − 0.9𝑥 − 1

𝑓 ′′′ (𝑥) = −2.4𝑥 − 0.9

𝑓 𝑖𝑣 (𝑥) = −2.4

𝑓 𝑣 (𝑥) = 0 Hence, Rn =0

𝑅4=𝑓 𝑛+1 ()

𝑛+1 !ℎ𝑛+1=

𝑓 𝑖𝑣 ()

5 !ℎ5 =

0

5!ℎ5 = 0


Example 4.1 – Successive Approximation

Here are the successive approximations for x=0 and

h=1 of 𝑓(𝑥) = −0.1𝑥4 − 0.15𝑥3 − 0.5𝑥2 − 0.25𝑥 + 1.2

𝑓 ′(0) = −0.4𝑥3 − 0.45𝑥2 − 𝑥 − 0.25 = −.25

𝑓 ′′ (0) = −1.2𝑥2 − 0.9𝑥 − 1 = −1

𝑓 ′′′ (0) = −2.4𝑥 − 0.9 = −0.9

𝑓 𝑖𝑣 (0) = −2.4

𝑓 𝑣 (0) = 0

𝑓0 1 = 𝑓 0 = 𝟏. 𝟐 𝑓1 1 = 𝑓0 1 + 𝑓′ 0 (1) = 1.2 − 0.25 1 = 𝟎. 𝟗𝟓

𝑓2 1 = 𝑓1 1 +𝑓′′(0)

2!(1)2= 0.95 −

1

2= 𝟎. 𝟒𝟓

𝑓3 1 = 𝑓2 1 +𝑓′′′ 0

3!(1)3= 0.45 −

0.9

6= 𝟎. 𝟑

𝑓4 1 = 𝑓3 1 +𝑓 𝑖𝑣 (0)

4!(1)4= 0.3 −

2.4

24= 𝟎. 𝟐 = 𝒇(𝟏)

𝑓0 1 = 𝑓(0)

𝑓𝑛 1 = 𝑓𝑛−1 1 +𝑓 𝑛 (0)

𝑛!ℎ𝑛


Order n f(p /3) t

0 cos(x) = 0.70710678 0.70710678 41.4214%

1 -sin(x) = -0.70710678 0.52198666 4.3973%

2 -cos(x) = -0.70710678 0.49775449 0.4491%

3 sin(x) = 0.70710678 0.49986915 0.0262%

4 cos(x) = 0.70710678 0.50000755 0.0015%

5 -sin(x) = -0.70710678 0.50000030 0.0001%

6 -cos(x) = -0.70710678 0.49999999 0.0000%

f[n]

(p /4)

Example 4.2 Infinite Taylor Series Approximation

Use Taylor Series expansions with n=0 through n=6 to approximate f(x) = cos(pi/3) on the basis of f(x) and derivatives at pi/4 (h=pi/3-i/4)


4.1.1 – The Remainder for the Taylor Series Expansion

Suppose we truncate the Taylor Series expansion

after the first term (zero-order approximation):

𝑓(𝑥 + ℎ) = 𝑓(𝑥) + 𝑅0

Where the 0th-order approximation error is give by:.

𝑅0 = (𝑥 − 𝑡)𝑛

𝑛!𝑓 𝑛+1 (𝑡)𝑑𝑡 =

𝑓 𝑛+1 ()

(𝑛 + 1)!ℎ𝑛+1 = 𝑓 ′()ℎ

𝑥+ℎ

𝑡=𝑥

Where is some value between x and x+h



R0 is the slope to the exact prediction:

Figure 4.2 Zero-order Taylor Series Prediction



R0 is the slope to the exact prediction:

Figure 4.3 Derivative Mean-Value theorem


4.1.2 Using the Taylor Series to Estimate Truncation Error

The remainder Rn is an exact measure of the error of truncation at term n of a Taylor Series

Even though Rn usually cannot be calculated exactly, the order of Rn is hn [ie O(hn )]

This allows us to estimate with good accuracy the error

In iterative procedures this is the stopping criterion.


Figure 4.4 In creasing non-linearity of a function requires higher and higher Taylor Series Approximations


Figure 4.5 Log-log plot of R1 vs h showing that as h gets smaller R1 decreases with h2 O(h2)


4.1.3 Numerical Differentiation

This will be covered in much more detail in Chapters 23 and 24.

Here we will introduce the three approximations typically used for a derivative o The forward difference o The backward difference o The centered difference

We will also look briefly at second derivatives.



The first forward difference approximation.

𝑓 ′(𝑥𝑖) =𝑓(𝑥𝑖+1) − 𝑓(𝑥𝑖)

ℎ+ 𝑂(ℎ) =

∆𝑓𝑖ℎ

+ 𝑂(ℎ)

The first backward difference approximation

𝑓 ′(𝑥𝑖) =𝑓(𝑥𝑖) − 𝑓(𝑥𝑖−1)

ℎ+ 𝑂(ℎ) =

∇𝑓𝑖ℎ

+ 𝑂(ℎ)



The first centered difference approximation.

𝑓 ′(𝑥𝑖) =𝑓(𝑥𝑖+1) − 𝑓(𝑥𝑖−1)

2ℎ+ 𝑂(ℎ2)

The second derivative approximation

𝑓 ′(𝑥𝑖) =𝑓(𝑥𝑖+1) − 2𝑓(𝑥𝑖) + 𝑓(𝑥𝑖−1)

ℎ2+ 𝑂(ℎ2)


f(x)=Ax4 + Bx3 + Cx2 +Dx + E f '(0.5) = -0.9125

A = -0.1 For h=0.5 error

B = -0.15 Df/h = -1.45 58.90%

C = -0.5 f/h = -0.55 39.73%

D = -0.25 cf/h = -1 9.59%

E = 1.2

x f(x)

0.00 1.20000000 For h=0.5 error reduction

0.25 1.10351563 Df/h = -1.15469 26.54% 45.06%

0.50 0.92500000 f/h = -0.71406 21.75% 54.74%

0.75 0.63632813 cf/h = -0.93438 2.40% 25.00%

1.00 0.20000000

Example 4.4 Pages 92-93 of the text


4.2 – Error Propagation

The purpose of this section is to investigate how errors propagate through mathematical functions.

For example, if we multiply two numbers together, each with a known error, what is the error of the

product?

We will look at error propagation for

o Single-variable functions

o Multi-variable functions o Stability and condition


4.2.1 – Single-Variable Functions

Assume we have a function f(x) that is

dependent on a single variable x.

We have a known value 𝑥 that is an

approximation to x

Then the error in using 𝑓(𝑥 ) rather than f(x) is

given by:

∆𝑓(𝑥 ) = 𝑓 ′(𝑥 ) ∆𝑥


4.2.1 – Single-Variable Functions

Then the error in using 𝑓(𝑥 ) rather than f(x) is given by: ∆𝑓(𝑥 ) = 𝑓 ′(𝑥 ) ∆𝑥


Example 4.5

Given 𝑥 = 2.5 with D𝑥 = 0.01, estimate the

resulting error in the function f(x) = x3.

Solution: We have f’(x) = 3x2.

Hence, f’(𝑥 ) = 3(2.5)2 = 18.75. Therefore:

∆𝑓(𝑥 ) = 𝑓 ′(𝑥 ) ∆𝑥

∆𝑓(𝑥 ) = 18.75 (0.01) = 0.1875

Predict: (2.5)3 – 0.1875 x3 (2.5)3 + 0.1875

15,4375 x3 15.8125


4.2.2 – Multi-Variable Functions

The concept of 4.2.1 can be extended to multi-variable functions:

∆𝑓(𝑥 1, 𝑥 2, ⋯𝑥 𝑛) ≅ 𝜕𝑓

𝜕𝑥1 ∆𝑥1 +

𝜕𝑓

𝜕𝑥2 ∆𝑥2 + ⋯ +

𝜕𝑓

𝜕𝑥𝑛 ∆𝑥𝑛

See Example 4.6 pp. 96-97.


4.2.3 – Stability and Condition

The condition of a mathematical problem relates to its sensitivity to changes in input values.

We say that a computation is numerically

unstable if the uncertainty of the input values is

grossly magnified by the numerical method.

A Condition Number can be defined by:

𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑁𝑢𝑚𝑏𝑒𝑟 =𝑥 𝑓 ′(𝑥 )

𝑓(𝑥 )


4.3 – Total Numerical Error

The Total Numerical Error is the summation of the truncation and round-off (or chopping) errors.

Round-off and chopping errors increase with subtractive cancellation and increased number of

calculations.

Truncation errors decrease with smaller step size,

but smaller step sizes increase the number of

calculations. Hence: There is a trade-off!


4.3 – Total Numerical Error

Truncation errors decrease with smaller step size, but smaller step sizes increase the number of

calculations. Hence: There is a trade-off!

Figure 4.8 Trade-off between Truncation and Round-Off errors.


Example 4.8 Truncation errors decrease with smaller step size,

but smaller step sizes increase the number of calculations. Hence: There is a trade-off!

Figure 4.9 Trade-off between Truncation and Round-Off errors in Example 4.8.


4.4 – Blunders, Formulation Errors and Uncertainty

Programming blunders are often a cause of

error that is difficult to eliminate.

Modeling Errors (also called Formulation

Errors) are also difficult to eliminate.

Data Uncertainty also causes errors, but we

do have some measures for this that can be very helpful in at least accounting for these

types of errors.

1/24/2011 ENGR-516 Prof. Soderstrand Spring 2011

End of Tonight’s Lecture

Any questions?

Homework for next week: 3.1, 3.3, 3.6

(note: use e-19 rather than e-5), 3.13, 4.1

and 4.23

Quiz for this week is NOW online –

Complete before class next week

Quiz for next week will be online next

Monday. 88

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Lecture 2 ENGR-516 Spring 2011

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