1. Recall our hypothetical computer Marc-32
Sign: 1 bit
Mantissa: 23 bits
Exponent: 8 bits
Normalized floating point representation
0 𝒙= 𝑞  2𝒎 (1𝑞2,−126𝑚127)
Unit roundoff error:  2𝟐𝟒
Floating point machine number 𝒇𝒍 𝒙
𝒇𝒍 𝒙 =𝒙 𝟏+ ||

2. Example 1 (P46) What’s the binary form of x=2/3
Example 1 (P46) What’s the binary form of x=2/3? What are two nearby machine numbers x- and x+ in the Marc-32? Which one is taken to be fl(x)? What are the absolute roundoff error and relative roundoff error in representing x by fl(x)?

3. Solution. First, we write 2/3 in the binary form
2/3 = (0.a1a2a3…) (1)
where ai’s are either 0 or 1.
We multiply by 2 for both sides to obtain
4/3 = (a1.a2a3…)2
Thus, we get a1=1 by taking the integer part of both sides. So,
1/3 = 4/3  1 = (a1.a2a3…)2  1= (0.a2a3a4…)2
2/3 = (a2.a3a4a5…) (2)

4. x? = (1.0101…011)2  2-1 (by rounding up)
From (1)-(2), we have
1=a1=a3=a5=a7=…
0=a2=a4=a6=a8=…
Thus,
x = 2/3 = ( …)2 = ( …)2  2-1
In Marc-32, the two nearby machine numbers are
x? = (1.0101…010)2  (by chopping)
x? = (1.0101…011)2  (by rounding up)
23 bits
Recall: x< x < x+

5. So, fl(x)=? Next, x  x = (1.01010…)2  2-24  2-1
= ( …)2  2-23  2-1
= 2/3  2-24
x x = (x x )  (x  x)
= 2-24  2/3  2-24 = 1/3  2-24
So, fl(x)=?

6. The absolute roundoff error
|fl(x)  x| = 1/3  2-24
The relative roundoff error
|fl(x)  x| |x| = 1/3  2−24 2/3 =2−25
Check your textbook (P47) for the decimal form of x, x and x  x

8. Stored as machine numbers fl(a),fl(b),… Rounded off (舍入)
Input numbers a,b,c,…
Normalized (标准化)
Stored as machine numbers fl(a),fl(b),…
Rounded off (舍入)
Do one arithmetic operation/calculation
Obtain a number (result) e.g. fl(a)fl(b)

10. The computer with 5 decimal digits stores those results in rounded form as
The relative errors are respectively

11. The computer with 5 decimal digits stores those results in rounded form as
The relative errors are respectively

12. Denote  one of the four basic arithmetic operations:   , 
Assume x,y are machine numbers, then
there is some constant  s.t.
fl(xy) = [xy] (1+)
where ||; here,  can be taken to be the unit roundoff error for the machine. In Marc-32, =2-24.
Q: How to compute xy if x,y are not machine numbers?

13. If x,y are not machine numbers, then
there is still some constant  s.t.
fl(x) = x (1+1)
fl(y) = y (1+2)
fl(xy) = fl(fl(x)fl(y))
= (fl(x)fl(y)) (1+3)
= [(x(1+1))  (y(1+2))](1+3)
= (xy)(1+1+2+12) (1+3)  xy
where |1|,|2|,|3|; still,  can be taken to be the unit roundoff error for the machine.

14. Q: How about compand arithmetic operations?
Assume x,y,z  A={machine numbers of Marc-32}.
fl(x(y+z))
= [x fl(y+z)] (1+1) |1| 2-24
= [x (y+z) (1+2)] (1+1) |2| 2-24
= x (y+z) (1+2+1 +12)
 x (y+z) (1+2+1)
= x (y+z) (1+3) |3| ?
Here 3=2+1

15. Exercise.
Find fl(x(y+z)) for
x, y, z  A={machine numbers of Marc-32}.

16. Theorem on Relative Roundoff Error in Adding

17. 2.2 Absolute & Relative Errors: Loss of Significance/Precision
Assume a real number 𝑥 is approximated by another number 𝑥 ∗ , the error is 𝑥−𝑥 ∗ .
The absolute error
|𝑥−𝑥 ∗ |
The relative error
|𝑥−𝑥 ∗ | |𝑥|

18. The relative error involved in representing a real number 𝑥 by a nearby floating-point machine number fl(𝑥) is bounded by the unit roundoff error 
|𝒙−𝒇𝒍(𝒙)| |𝒙| 
Roundoff errors are inevitable & difficult to control.

19. Loss of Significance
The subject of numerical analysis is largely involved in understanding and controlling errors of various kinds.

20. For example,
𝑥= ,𝑦=
𝑥−𝑦=
If this calculation were to be performed in a decimal computer having a five-digit mantissa, we would have
𝑓𝑙 𝑥 = , 𝑓𝑙(𝑦)=
𝑓𝑙(𝑥)−𝑓𝑙(𝑦)=
The relative error
|𝒙−𝒚−[𝒇𝒍 𝒙 −𝒇𝒍 𝒚 ]| |𝒙−𝒚| 𝟒%

21. Loss of Significance
The result is usually stored as a normalized floating-point number, i.e.,
𝒇𝒍 𝒙 −𝒇𝒍 𝒚 =𝟎.𝟎𝟎𝟎𝟏𝟑
=𝟎.𝟏𝟑𝟎𝟎𝟎𝟏𝟎𝟑
The added three 0’s in above do NOT represent additional accuracy, i.e., those three additional 0’s are NOT significant numbers (有效数字).

22. Subtraction of Nearly Equal Quantities
Example 1 The assignment statement
𝑦  𝑥 2 +1 −1
can cause loss of significance for small values of 𝑥. How to avoid this trouble?
Solution. The statement can be replaced by
𝑦  𝑥 2 /( 𝑥 )
in programming to avoid such trouble.

23. Ex. 2：
求根（保留小数点后10位）
e-4

24. Loss of Precision Theorem 1(P57) Theorem on Loss of Precision
If 𝑥 and 𝑦 are positive normalized floating-point binary machine numbers such that 𝑥𝑦 and
2 −𝑞  1− 𝑦 𝑥  2 −𝑝
then at most 𝑞 and at least 𝑝 significant binary bits are lost in the subtraction 𝑥𝑦.

25. Proof. Only prove the lower bound and leave the upper bound as your after-class exercise.
The normalized binary floating-point forms for 𝑥,𝑦 are
𝒙=𝒓  𝟐𝒏 , 𝒚=𝒔  𝟐𝒎 , ( 1 2 𝑟,𝑠1)
Since 𝑥𝑦, the computer may have to shift 𝑦 so that 𝑦 has the same exponent as 𝑥 before performing 𝑥𝑦.
So, we must write 𝑦 as 𝑦= 𝒔  𝟐𝒎𝒏  𝟐𝒎
and then
𝑥𝑦=(𝒓 𝒔  𝟐𝒎𝒏) 𝟐𝒏

26. 𝑟−𝒔𝟐𝒎𝒏=𝒓 𝟏− 𝒔𝟐𝒎 𝒓𝟐𝒏 =𝒓 𝟏− 𝑦 𝑥 𝟏− 𝑦 𝑥  2 −𝑝
By assumption, we have
𝑟−𝒔𝟐𝒎𝒏=𝒓 𝟏− 𝒔𝟐𝒎 𝒓𝟐𝒏 =𝒓 𝟏− 𝑦 𝑥 𝟏− 𝑦 𝑥  2 −𝑝
WLOG, assume the mantissa in the computer has 𝑝+𝑘 digits (𝑘1), then
𝒓−𝒔𝟐𝒎𝒏= 𝟎.𝟎𝟎𝟎𝒂𝟏𝒂𝟐𝒂𝒌 𝟐 𝑝
The normalized floating point form of 𝑥𝑦 is
𝑥−𝑦= 𝟎. 𝒂𝟏𝒂𝟐𝒂𝟑𝒂𝒌𝟎𝟎 𝟐 𝟐𝒏𝒑 𝑖𝑓 𝒂𝟏 0 𝟎. 𝒂𝟐𝒂𝟑𝒂𝒌𝟎𝟎𝟎 𝟐 𝟐𝒏𝒑𝟏 𝑖𝑓 𝒂𝟏=0,𝒂𝟐 0  

27. i.e., a shift of at least 𝑝 bits to the left is required; meanwhile, at least 𝑝 spurious 0’s are attached to the right end of the mantissa, which means that at least 𝑝 bits of precision have been lost.

28. 𝑦𝑥−sin⁡(𝑥) Example 3. Consider the assignment statement
This calculation involves a loss of significance for small values of 𝑥. How to avoid this trouble?

29. Solution. By the Taylor series for sin(𝑥), we have 𝑦=𝑥− sin 𝑥
=𝑥−(𝑥− 𝑥3 3! + 𝑥5 5! − 𝑥7 7! +)
= 𝑥3 3! − 𝑥5 5! + 𝑥7 7! − 𝑥9 9! 
If 𝑥 is near 0, a truncated series can be used, e.g.,
𝑦 (𝑥 3 /6)(1− (𝑥 2 /20)(1− (𝑥 2 /42)(1− 𝑥 2 /72)))
Note that both assignment statements may be used for a wide range of values of 𝑥.

30. Homework & Programming
Check the course’s webpage for
Homework # Due Thursday, 9. 29
Programming #1 Due Thursday, 9. 29