Transcript Document

MA/CS 375
Fall 2002
Lecture 6
MA/CS375 Fall 2002
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Exercise: Blending Two Images
• Find one partner to work with (i.e. your neighbor)
• Write a script which reads in two images.
• Use the min function to find the minimum number of rows for
both pictures (MINrows)
• Use the min function to find the minimum number of columns
for both pictures (MINcolumns)
• Crop the images using
> pic1 = pic1(1:MINrows,1:MINcols,1:3)
> pic2 = pic2(1:MINrows,1:MINcols,1:3)
• Create a third picture (pic3) which is the average of pic1 and
pic2
• Plot pic1,pic2,pic3 in the same figure using subplot
•MA/CS375
Put Fall
your
names on the title, print it out and hand it in.
2002
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Finite Precision Effects
• Recall that we are computing in a finite
precision computing paradigm.
• i.e. all non integer numbers are
represented by a finite approximation.
• e.g. 1/2 is represented exactly, but 1/3
is not.
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Non-Special IEEE Floating Point
Double Precision
• A double precision (Matlab default) number is
represented by a 64 bit binary number:
( 1)  1.m  2
s
e 1023
• Here s is the sign bit
• m is the mantissa (length to be determined)
• e is the exponent and 0 < e < 2047
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A Convergent Binary
Representation of Any Number
Between 0 and 1
n 
1
m   mn n where mn  1 or 0
2
n 1
a similar representation in base 10:
n 
1
d   d n n where d n  0,1,2,3,4,5,6,7,8 or 9
10
n 1
Volunteer ?
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MA/CS375 Fall 2002
Finite Binary Approximations of a
Real Number
n N
1
m   mn n +TN where mn  1 or 0
2
n 1
1
We can easily show that TN is bounded as: TN  N
2
(think of Zeno’s paradox)
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Review
Definition of a Derivative
• Suppose we are given a function: f :

– where f is smooth and well defined at
every point on the real numbers.
– we can define the derivative of f at a point
x as:
df  x 
 f  x    f  x 
 lim 

dx

 0 

– later on today we will try to use this
definition (and see what goes wrong  )
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Monster Functions
Due to some of these finite precision effects
there are some odd behaviors of Matlab
documented in:
“Numerical Monsters”, Christopher Essex,
Matt Davison and Christian Schulzky.
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Team Tasks
We are going to investigate several examples
from:
“Numerical Monsters”, Christopher Essex,
Matt Davison and Christian Schulzky.
A.K.A “When Good Formulas Go Bad in Matlab”
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Teams
• Form teams of four
• Choose one of the following monster
exercises
• Remember to put your names in the
titles
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Monster #1
• Consider:
• What should its behavior be as:
f  x
1 x 1


x 0
x
• Use subplots for the following 2 plots
• Plot this function at 1000 points in:
x   1,1
• Plot this function at 1000 points in:
 4 ,4 
• Label everything nicely, include your name in the title.
• In a text box explain what is going on, print it out and hand it
in
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Monster #2
• Consider:
f  x   e log(1  e )
x
x
• What should its behavior be as:
x
• Plot this function at 1000 points in: x  0,50
• Explain what is going on in a text box, label
everything, print it out and hand it in.
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Monster #3
• Consider:
f  x   2 log 2 (1  2 )
x
x
• What should its behavior be as:
x
• Plot this function at 1000 points in:
x  0,60
• Explain what is going on in a text box and in
particular what happens at x=54, label everything,
print it out and hand it in.
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Monster #4
 sin  x     sin  x  
g  x   cos  x   




• What should its behavior be as:
 0
• Consider:
• Plot four subplots of the function at 1000 points in:
x  0,2 *   for   1e  4,1e  7,1e  12,1e  15
• Now fix x=0.5 and plot this as a function of
  0.1,0.01,0.001,...,1e  20

for
• Explain what is going on, print out and hand in.
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Explanations of Team Examples
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Recall
Monster #1
• Consider:
f  x
1 x 1


x
• What should its behavior be as: x  0
• Plot this function at 1000 points in:
• Plot this function at 1000 points in:
x   1,1
 4 ,4 
• Explain what is going on.
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Monster #1
((large+small)-large)/small
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Monster #1
((large+small)-large)/small
when we zoom in we see that the large+small operation is introducing
order eps errors which we then divide with eps to get O(1) errors !.
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Monster #1
((large+small)-large)/small
Each stripe is a region
where 1+ x is a constant
(think about the gaps between
numbers in finite precision)
Then we divide by x and the
stripes look like hyperbola.
The formula looks like
(c-1)/x with a new c for each
stripe.
when we zoom in we see that the large+small operation is introducing
order eps errors which we then divide with eps to get O(1) errors !.
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Recall
Monster #2
• Consider:
f  x   e log(1  e )
x
x
• What should its behavior be as:
x
• Plot this function at 1000 points in: x  0,50
• Explain what is going on in a text box, label
everything, print it out and hand it in.
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Limit of

f  x   e log(1  e )
lim f  x   lim e log 1  e
x 
x 
x
x
x

x
 log 1  e  x  
 lim 


x
x  

e


 e  x 
 1  e x 
 lim 
by l'Hopital's rule


x
x 
  e 


 1 
 lim 
1


x
x  1  e


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Monster #2
(finite precision effects from large*small)
f  x   1 as x  
As x increases past 30 we see
that f deviates from 1 !!
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Monster #2 cont
(finite precision effects from large*small)
As x increases past ~=36 we
see that f drops to 0 !!
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Recall
Monster #3
• Consider:
f  x   2 log 2 (1  2 )
x
• What should its behavior be as:
x
x
• Plot this function at 1000 points in: x  0,60
• Explain what is going on. What happens at
x=54?
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Monster 3
(finite precision large*small with binary stripes)
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Monster 3
(finite precision large*small with binary stripes)
As we require more than 52 bits to represent
1+2^(-x) we see that the log term drops to 0.
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Recall
Monster #4
 sin  x     sin  x  
g  x   cos  x   




• What should its behavior be as:
 0
• Consider:
• Plot four subplots of the function at 1000 points in:
x  0,2 *   for   1e  4,1e  7,1e  12,1e  15
• Now fix x=0.5 and plot this as a function of
  0.1,0.01,0.001,...,1e  20

for
• Explain what is going on, print out and hand in.
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Monster 4 cont
Behavior as delta  0 :
 sin  x     sin  x  
 sin  x  cos    cos( x )sin    sin  x  
lim 
 lim 


 0
 0






 sin  x   cos    1  cos( x )sin   
 lim 

 0 



 cos  x 
or if you are feeling lazy use the definition of derivative, and remember:
d(sin(x))/dx = cos(x)
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Monster 4 cont
(parameter differentiation, delta=1e-4)
OK
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Monster 4 cont
(parameter differentiation, delta=1e-7)
OK
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Monster 4 cont
(parameter differentiation, delta=1e-12)
Worse
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Monster 4 cont
(parameter differentiation, delta=1e-15)
Bad
When we make the delta around about machine precision we see
O(1) errors !.
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Monster 4 cont
(numerical instablitiy of parameter differentiation)
As delta gets smaller we see that the approximation improves, until delta ~= 1e-8 when
it gets worse and eventually the approximate derivate becomes zero.
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Approximate Explanation of
Monster #4
f  x     f  x    f '  x   O  2 
1) Taylor’s thm:
yˆ  xˆ  ˆ
 x    O ( )
2) Round off errors
3) Round off in computation
of f and x+delta
fˆ  yˆ   f  yˆ   O ( )
 f  x     O ( )
fˆ  yˆ   fˆ  xˆ   f  y   f  x   O  
4) Put this together:
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  f '  x   O  2   O   
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ˆf  yˆ   fˆ  xˆ    f '  x   O  2   O  
 fˆ  yˆ   fˆ  xˆ  
 

 O    O  









i.e. for 

or equivalently 

  10
8
approximation error decreases as delta
decreasesize.
BUT for 
MA/CS375 Fall 2002
  10
8
round off dominates!.
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Summary
• Ok – so in the far limits of the range of finite
precision really dodgy things happen.
• But recall, the formula for the derivative of sine
worked pretty well for a large range of numbers.
• Try to avoid working with two widely separated
numbers.
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Summary of Lecture 6
• Today’s lecture was all about the way Matlab
(and in fact almost all languages) store
representations of real numbers
• We found out:
– because the set of possible double precision
numbers only covers a discrete subset of points
on the real line there are finite gaps between
consecutive numbers
– If A is “large” and B is very “small” the sum A+B
actually results in A+B ~= A
– and so on..
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