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Quiz #1
CSC2800 (Spring Term 2007)
Full score: 10 % + 1% bonus
Name:_______________________________ (Student ID: ________________________)
Problem 1:
A fictional machine represents non-zero floating point numbers in the following way:
  0.a1a2 a3a4  10 e
a1  0,
0  a1 , a2 , a3 , a 4  10 , and  5  e  5
That is, the mantissa can only support 4 decimal digits. In addition, numbers that cannot be represented exactly are
rounded to the nearest representation.
[This is the same fictional machine you encountered in your assignment #1]
(a) What is the largest representable positive number on this machine?
Answer:  0.9999  10
[1%]
5
(b) What is the smallest representable positive number on this machine?
[1%]
Answer:  0.1000  10 4
(c) Suppose x and y are two representable non-zero numbers of this machine. Find an example such that x / y (x
divide by y) overflow.
[1%]
Answer:
x  0.9999  10 5 , y  0.1 (or any number less than 1)
0.9999  10 5 / 0.1  0.9999  10 6 ( the largest representa ble number, and thus results in overflow)
(c) Give an example to show that adding two numbers can result in underflow on this machine.
[1%]
Answer:
Let x  0.1001  10 5 , y  0.1000  10 5
Then x  y  0.0001  10 5 ( the smallest representa ble positive number, and thus results in underflow)
Problem 2:
Suppose in computing x2 + y, we begin with x and y with some errors. Let εx and εy be the errors introduced to x
and y respectively. That is, xA = x + εx and yA = y + εy. Assuming there is no rounding error introduced during
calculation, what is the total errors introduced to the final result?
[2%]
Answer:
Total error
 ( x 2  y )  ( x A2  y A )
 ( x 2  y )  (( x   x ) 2  y   y )
 x 2  y  x 2  2 x x   x2  y   y
 2 x x   x2   y
Problem 3:
Given f(x) = x4 + 2x3 + 2x + 5
(a) How can we compute f(x) efficiently and at the same time minimize the effect of rounding errors.
[1%]
Answer:
Evaluate f(x) as ((x + 2)x2 + 2)x + 5
(b) What is the minimum number of terms in the Taylor expansion of f(x) at 1000 is needed to represent f(x)
with no truncation errors?
[1%]
Answer: 5 terms.
Problem 4:
Given f(x) = ex – 1.
(a) What values of x could lead to large error in the calculated result due to subtractive cancellation?
[1%]
Answer: When x is close to 0 (which makes ex is close to 1).
(b) How should we compute f(x) for the values of x you specified in part (a)?
Note: The Taylor series of ex at 0 is 1  x 
[1%]
x2 x3
xn

 ... 
 ...
2! 3!
n!
Answer:
x 2 x3
xn
x2
For x  0, calculate f ( x) as f ( x)  e  1  (1  x    ... 
 ...)  1  x  
2! 3!
n!
2!
2
3
x
x
 ( x    ... 
y
1
1 e
2! 3!
For x  0, set y   x and calculate f ( x) as f ( y )  y  1  y 
e
e
ey
x
x3
xn
 ... 
 ...
3!
n!
xn
 ...)
n!
Bonus:
Prove or explain why your answer for 3(b) is indeed the minimum number of terms needed.
[1%]
Answer:
f ( n 1) (c)
The truncation error of the Taylor series of f(x) after including the first n+1 terms is
( x  1000)n 1 for
(n  1)!
( 4)
( n)
some c between 1000 and x. Since f ( x)  24  0 and f ( x)  0 for n ≥ 5, the smallest n that results in
zero truncation error is 5.