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CSC336 Tutorial 2 – Computer arithmetic
创建日期：2024-04-26 10:01:58
标签： CSC336
Computer arithmetic

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CSC336 Tutorial 2 – Computer arithmetic

QUESTION 1 Find the positive numbers inR2(4, 3) assuming normalized mantissa.
ANSWER: The numbers in R2(4, 3) are of the form f × 2e, where f the mantissa
and e the exponent. The table below indicates all possible positive (and normalized)
values of the mantissa f , all possible values for the exponent e, and the numbers of
the form f × 2e.
Notes:
1. (1) does not need to be stored since we assume normalized mantissa.
2. The 4th bit of the mantissa is used for the sign. R2(4, 3) also includes the re-
spective negative numbers, as well as 0, which is handled as a special number.
3. Figure 1 shows the position of the numbers on the axis. Numbers get denser
towards zero, while numbers with the same exponent are uniform.
Tut2 – Computer arithmetic 1 c© C. Christara, 2012-22
Table 1: f × 2e
e −(11)2 −(10)2 −(01)2 −(00)2 (01)2 (10)2 (11)2
-3 -2 -1 0 1 2 3
2e 1/8 1/4 1/2 1 2 4 8
f
(.(1)000)2 8/16 1/16 1/8 1/4 1/2 1 2 4
(.(1)001)2 9/16 9/128 9/64 9/32 9/16 9/8 9/4 9/2
(.(1)010)2 10/16 5/64 5/32 5/16 5/8 5/4 5/2 5
(.(1)011)2 11/16 11/128 11/64 11/32 11/16 11/8 11/4 11/2
(.(1)100)2 12/16 3/32 3/16 3/8 3/4 3/2 3 6
(.(1)101)2 13/16 13/128 13/64 13/32 13/16 13/8 13/4 13/2
(.(1)110)2 14/16 7/64 7/32 7/16 7/8 7/4 7/2 7
(.(1)111)2 15/16 15/128 15/64 15/32 15/16 15/8 15/4 15/2
Tut2 – Computer arithmetic 2 c© C. Christara, 2012-22
0 1 2 3 4 5 6 7 8
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Figure 1: Positive numbers inR2(4, 3)
Tut2 – Computer arithmetic 3 c© C. Christara, 2012-22
QUESTION 2 Let x be a number and let fl(x) be its floating-point representation
with t-digit mantissa, in base b. Prove that the following bounds for the relative
round-off error δ = (x− fl(x))/x hold:
(i) chopping and normalized: |δ| ≤ b1−t.
(ii) traditional rounding and normalized: |δ| ≤ 12b1−t.
PROOF:
(i) Case x > 0, i.e. x ≥ fl(x), therefore, |δ| = δ = (x− fl(x))/x
|δ| = .d1d2 . . . dtdt+1 . . .× b
e − .d1d2 . . . dt × be
.d1d2 . . . dtdt+1 . . .× be
=
.dt+1dt+2 . . .× b−t
.d1d2 . . .
≤ .dt+1 . . .× b
−t
.100 . . .
[Note: .100. . . is the smallest normalized number]
≤ .aa . . .× b
−t
.1
where a = b− 1
=
1× b−t
.1
(∗)
= b1−t
Tut2 – Computer arithmetic 4 c© C. Christara, 2012-22
(ii) Case x > 0, dt+1 < b/2, i.e. x ≥ fl(x), therefore, |δ| = δ = (x− fl(x))/x
|δ| = .d1d2 . . . dtdt+1 . . .× b
e − .d1d2 . . . dt × be
.d1d2 . . . dtdt+1 . . .× be
=
.dt+1dt+2 . . .× b−t
.d1d2 . . .
≤ .caa . . .× b
−t
.1
where c = b/2− 1
=
1
2
b1−t (∗∗)
Case x > 0, dt+1 ≥ b/2 i.e. x < fl(x), therefore, |δ| = −δ = −(x− fl(x))/x
|δ| = fl(x)− x
x
=
.d1d2 . . . (dt + 1)× be − .d1d2 . . . dtdt+1 . . .× be
.d1d2 . . . dtdt+1 . . .× be
=
b−t + .d1d2 . . . dt − .d1d2 . . . dtdt+1 . . .
.d1d2 . . .
≤ b
−t + .d1d2 . . . dt − .d1d2 . . . dtc
.d1d2 . . .
where c = b/2
Tut2 – Computer arithmetic 5 c© C. Christara, 2012-22
=
b−t − cb−t−1
.d1d2 . . .
≤ b
−t − cb−t−1
.1
= b1−t − b
2
b−t
= b1−t − 1
2
b1−t
=
1
2
b1−t
The proof for x < 0 is similar.
Notes: Σ = α + αw + αw2 + . . . + αwν = α(w
ν+1−1)
w−1 ,Σ[1,∞),w<1 =
α
1−w
Therefore:
(*) (.aa . . .) =
∑∞
i=1 ab
−i = a b
−1
1−b−1 = (b− 1) 1b−1 = 1
(**) by (*), .0aa . . . = 1b(.aa . . .) =
1
b = b
−1
.caa . . . = .0aa . . . + .c = b−1 + (b
2
− 1)b−1 = b
2
b−1 = 1
2
Tut2 – Computer arithmetic 6 c© C. Christara, 2012-22
QUESTION 3 Find ǫmach for a floating-point system with t binary digits precision and
(i) traditional rounding or (ii) proper rounding. (Answer in terms of t.)
ANSWER: Recall that ǫmach is the smallest (non-normalised) floating-point number
of the form b−i, for i some positive integer, with the property 1 + ǫmach > 1. The
floating-point representation of “1” (with binary mantissa) is 1 = 0.100 · · · 0× 21. We
construct ǫmach with the smallest possible mantissa and the smallest possible exponent,
so that 1 + ǫmach > 1 holds.
(i) traditional rounding: Consider the operations 1+2−t and 1+2−t−1 (mantissa shown
in binary):
0.100 · · · 000× 21 0.100 · · · 000× 21
+ 0.000 · · · 010× 21 + 0.000 · · · 001× 21
= 0.100 · · · 010× 21 = 0.100 · · · 001× 21
where we (temporarily) show t + 2 bits
in the mantissa.
Note that 0.100 · · · 010× 21 is rounded up to 0.100 · · · 100× 21 = 0.100 · · · 1× 21 > 1
(the t + 1st digit is b/2, so round-up), while 0.100 · · · 001 × 21 is rounded down to
0.100 · · · 0 × 21 = 1 (the t + 1st digit is 0, so less than b/2, so round-down). Thus,
ǫmach = 0.000 · · · 010× 21 = 0.000 · · · 1× 20 = 2−t.
(ii) proper rounding: Consider the operations 1 + 2−t+1 and 1 + 2−t (mantissa shown
Tut2 – Computer arithmetic 7 c© C. Christara, 2012-22
in binary):
0.100 · · · 000× 21 0.100 · · · 000× 21
+ 0.000 · · · 100× 21 + 0.000 · · · 010× 21
= 0.100 · · · 100× 21 = 0.100 · · · 010× 21
where we (temporarily) show t + 2 bits
in the mantissa.
Note that 0.100 · · · 100 × 21 = 0.100 · · · 1 × 21 > 1 (remains as is, no round-up or
down), while 0.100 · · · 010 × 21 is rounded down to 0.100 · · · 0 × 21 = 1 (the t + 1st
digit is b/2; if rounded-up we end with an odd digit, so it is rounded-down). Thus,
ǫmach = 0.000 · · · 100× 21 = 0.000 · · · 10× 20 = 2−t+1.
QUESTION 4 Find ǫmach for a binary floating-point system with proper rounding,
without knowing the number t of digits in the mantissa. (Essentially, find t.)
ANSWER: Consider the code (written in MATLAB for convenience)
x = 1; i = 0;
while (1+x > 1)
x = x/2; i = i+1;
end
me = 2*x; fprintf(’ machine epsilon = %22.15e t = %d\n’, me, i);
This code goes through the numbers x = 1, (0.1)2 = 2
−1, (0.01)2 = 2−2, . . ., up to the
point that 1 + x = 1, where x = 2−i, and outputs 2x, i.e. 2−i+1, which is the smallest
Tut2 – Computer arithmetic 8 c© C. Christara, 2012-22
x for which 1 + x > 1. Thus ǫmach = 2
−i+1. The code also outputs (the last) i.
In a previous discussion we showed that ǫmach for proper rounding is 2
−t+1, where t is
the number of digits in the mantissa. Thus we can deduce that the number t of digits
in the mantissa is i as output by the code.
If we run this code in MATLAB we get as output
machine epsilon = 2.220446049250313e-016 t = 53.
Note that this corresponds to double precision IEEE standard (as given in the notes).
In general, MATLAB uses double precision.
Note also, that MATLAB has a predefined variable eps which gives the value of the
machine epsilon, so we need not run the code to find it. We can, though, check that
MATLAB’s eps and me from the above code have the same value, 2−52.
In C, C++, fortran or other languages, we can run a similar code using either floats
(reals) or doubles, and find ǫmach for single or double precision, respectively. (If tradi-
tional rounding is used, we need to adjust the code accordingly.)
Tut2 – Computer arithmetic 9 c© C. Christara, 2012-22
QUESTION 5 A computer system uses decimal floating-point arithmetic with 3 digits
mantissa (plus one digit for the mantissa sign) and 1 digit exponent (plus one digit
for the exponent sign). All computer operations return the correctly rounded result
(i.e. the number closest to the correct answer in the representation being used). Com-
pute the results of the following arithmetic operations on the above computer system.
Indicate when overflow or underflow occurs. Let a = 0.697 and b = 0.699.
(336 + 1.42) + 0.1 = 337.42 + 0.1 → 337 + 0.1
= 337.1 → 337 = .337× 103
336 + (1.42 + 0.1) = 336 + 1.52 → 336 + 1.52
= 337.52 → 338 = .338× 103
336 · 107 = 336 · 107 → .336× 1010 overflow
−900 · 106 − 100 · 106 = −1000 · 106 → −.100× 1010 overflow
108 + 105 = 1001 · 105= .1001 · 109 → .100× 109 (= 108)
(a + b)/2 = (0.697 + 0.699)/2 = 1.396/2 → 1.40/2 = 0.7 = 0.700×
a+(b−a)/2 = 0.697+(0.699−0.697)/2 = 0.697 + 0.002/2 → 0.697+(0.2×10−2)/2
= 0.697+0.001=0.698 → 0.698× 100
Tut2 – Computer arithmetic 10 c© C. Christara, 2012-22
Notes:
For each individual mathematical operation, we first compute the correct mathemati-
cal result (indicated by the = sign), then show the computational result as represented
in the particular computer system given (indicated by the→ sign), by applying round-
ing (whenever needed).
In showing each computational result we take into account that the respective math-
ematical result is written by possibly shifting the mantissa digits and adjusting the
exponent accordingly (though, in some obvious cases, we do not explicitly give this
presentation of the result).
We can also use the fl notation to show the computed result for the above operations.
This notation is more formal, but possibly a bit cumbersome. For example, for the
operations (336 + 1.42) + 0.1, we have
fl(fl(fl(336) + fl(1.42)) + fl(0.1)) = fl(fl(336 + 1.42) + 0.1) representation
= fl(fl(337.42) + 0.1) operation
= fl(337 + 0.1) representation
= fl(337.1) operation
= 337 = .337× 103 representation
Tut2 – Computer arithmetic 11 c© C. Christara, 2012-22
Note that the exact mathematical result is 337.42 (which, if rounded, gives 337).
If we change 336 to 336.1, and do the operations (336.1 + 1.42) + 0.1, we have
fl(fl(fl(336.1) + fl(1.42)) + fl(0.1)) = fl(fl(336 + 1.42) + 0.1) representation
= fl(fl(337.42) + 0.1) operation
= fl(337 + 0.1) representation
= fl(337.1) operation
= 337 = .337× 103 representation
Note that the exact mathematical result is 337.52 (which, if rounded, gives 338).
By comparing the two examples ((336 + 1.42) + 0.1 and (336.1 + 1.42) + 0.1), we see
that it is not enough to do all operations mathematically, then take the floating-point
respresentation of the final result. We need to take the floating-point respresentation
of each individual number or result, before each operation takes place. In the (336 +
1.42) + 0.1 case, there is no initial representation error (no need to round), while, in
the (336.1 + 1.42) + 0.1 case, there is some initial rounding error.
Tut2 – Computer arithmetic 12 c© C. Christara, 2012-22
QUESTION 6 Examples of catastrophic cancellation and how to avoid it.
(a) Solve the quadratic equation x2−56x+1 = 0, using 5-decimal-digits floating-point
arithmetic.
x1 = 28 +
√
783 = 28 + 27.982 = 55.982 = 0.55982× 102
x2 = 28−
√
783 = 28− 27.982 = .018 = .18× 10−1
x1(exact) = 55.982137 . . .
x2(exact) = .0178628 . . . , x2 only 2 digits correct
The root x2 is only 2 digits correct due to cancellation between 28 and
√
783.
A better way to compute the root x2 is to first compute x1 (which does not suffer from
cancellation) and then let x2 = 1/x1 = .017863. Thus we have 5 digits correct.
(b) Consider computing y =
√
x + δ −√x, when |δ| ≪ x, x > 0.
Here we have again subtraction of nearly equal numbers. To avoid subtraction of
nearly equal numbers, compute y by
x + δ − x√
x + δ +
√
x
=
δ√
x + δ +
√
x
.
(c) Consider computing y = cos(x + δ)− cos(x), when |δ| ≪ x.
Again we have subtraction of nearly equal numbers. To avoid subtraction of nearly
equal numbers, compute y by y = −2 sin δ2 sin(x + δ2).
Tut2 – Computer arithmetic 13 c© C. Christara, 2012-22
QUESTION 7 Study the condition number of the function f(x) = 10
1−x2 .
The condition number of f(x) is κf = |xf
′(x)
f(x)
| = |
x −2x·10
(1−x2)2
10
1−x2
| = | −2x
2
1− x2|
When is κf large? We need to check the cases that the numerator in κf is large and
the denominator is small.
Study the case when the numerator is large, that is, |x| is large.
We have
−2x2
1− x2 →x→±∞=
−4x
−2x = 2 (by de l’Hospital’s rule), thus κf → 2 which is
small.
Study the case when the denominator is small, that is, |x| → 1.
We have
−2x2
1− x2 →|x|→1 ±∞
So κf →∞, thus the condition number will be very large if |x| → 1.
See Figure 2, where the solid line is κf = |−2x21−x2| versus x, and the dashed line is −2x
2
1−x2
versus x.
x = linspace(-2, 2, 1000);
kf = abs(2*x.ˆ2./(1-x.ˆ2));
kf1 = -2*x.ˆ2./(1-x.ˆ2);
plot(x, kf, ’r-’, x, kf1, ’g--’);
hax = gca; set(hax, ’FontSize’, 16, ’TickLength’, [0.02 0.05])
print -depsc testcondf.m1.eps
Tut2 – Computer arithmetic 14 c© C. Christara, 2012-22
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−1000
−800
−600
−400
−200
0
200
400
600
800
1000
Figure 2: Condition number of f(x) = 10
1−x2
Tut2 – Computer arithmetic 15 c© C. Christara, 2012-22
Now consider f(x) =
10
1− x2 again, and give examples of what it practically means to
have a large condition number when |x| → 1.
Let x = 1+ δ, for δ = −10−1, . . . ,−10−3, 10−3, . . . , 10−1. Let y = x+10−5. Compute
κf(x), and
f(x)− f(y)
f(x)
. In this way, we perturb x a little (absolute perturbation 10−5,
relative perturbation 10−5/x), and study the relative perturbation
f(x)− f(y)
f(x)
in f(x).
Theoretically we expect
f(x)−f(y)
f(x)
x−y
x
≈ κf(x). That is, when x is close to 1, since
κf(x) is large, for a small relative perturbation 10
−5/x in x, the relative perturbation
f(x)− f(y)
f(x)
is expected to be large.
Figure 3 as well as the output of the code verifies that
f(x)−f(y)
f(x)
x−y
x
behaves approximately
as κf . The solid line is κf(x) versus x, and the dashed line is
f(x)−f(y)
f(x)
x−y
x
versus x. Notice
that we study only the [1− 10−1, 1+10−1] range of x (but we expect similar behaviour
everywhere).
Tut2 – Computer arithmetic 16 c© C. Christara, 2012-22
d = [-10.ˆ[-1:-1:-3] 10.ˆ[-3:-1]]; e = 1e-5;
fprintf(’ errx errf rerrx rerrf rerrf/rerrx condf\n’);
x = 1+d; x2 = x.ˆ2;
y = x+e; y2 = y.ˆ2;
f = 10./(1-x2);
g = 10./(1-y2);
errx = x - y; rerrx = errx./x;
errf = f - g; rerrf = errf./f;
kf = abs(2*x2./(1-x2));
ratio = abs(rerrf./rerrx);
plot(1+d, kf, ’r-’, 1+d, ratio, ’b--’)
hax = gca; set(hax, ’FontSize’, 16, ’TickLength’, [0.02 0.05])
axis tight
print -depsc testcondf.m2.eps
for i = 1:length(d);
fprintf(’%9.2e %9.2e %9.2e %9.2e %9.2e %9.2e\n’, ...
errx(i), errf(i), rerrx(i), rerrf(i), rerrf(i)/rerrx(i), kf(i))
end
Tut2 – Computer arithmetic 17 c© C. Christara, 2012-22
errx errf rerrx rerrf rerrf/rerrx condf
-1.00e-05 -4.99e-03 -1.11e-05 -9.47e-05 8.53e+00 8.53e+00
-1.00e-05 -5.00e-01 -1.01e-05 -9.96e-04 9.86e+01 9.85e+01
-1.00e-05 -5.05e+01 -1.00e-05 -1.01e-02 1.01e+03 9.99e+02
-1.00e-05 -4.95e+01 -9.99e-06 9.91e-03 -9.92e+02 1.00e+03
-1.00e-05 -4.99e-01 -9.90e-06 1.00e-03 -1.01e+02 1.02e+02
-1.00e-05 -4.99e-03 -9.09e-06 1.05e-04 -1.15e+01 1.15e+01
Tut2 – Computer arithmetic 18 c© C. Christara, 2012-22
0.9 0.95 1 1.05 1.1
100
200
300
400
500
600
700
800
900
1000
Figure 3:
f(x)−f(y)
f(x)
x−y
x
≈ κf(x)
Tut2 – Computer arithmetic 19 c© C. Christara, 2012-22
QUESTION 8 Study three ways to approximate ex. Consider the Taylor’s series ex =
1+x+ x
2
2! + · · ·, and compute the above sum from left to right up to saturation (i.e. up
to the point where the sum does not change), using three techniques:
(a) compute the sum from left to right up to saturation as given,
(b) compute the sum from left to right up to saturation using the fact that each term
added is the previous term times x
k
, and
(c) do as in (b), but using the absolute value of x, then, if x ≥ 0 return the result as is,
while if x < 0 return the inverse of the result (taking into account that e−x = 1ex).
Compute and tabulate the error in each case and discuss advantages and disadvan-
tages of each approximation. Assume that the function exp(x) returns the exact value
of ex.
ANSWER: Here is the code for the three approximations to ex, the main script and
the results.
Tut2 – Computer arithmetic 20 c© C. Christara, 2012-22
% function [y, k] = exp0(x)
% eˆx ˜ 1 +x +xˆ2/2! +xˆ3/3! + ...
%
function [y, k] = exp0(x)
sump = 0;
sumn = 1;
k = 0;
while sump ˜= sumn
sump = sumn;
k = k+1;
term = xˆk/factorial(k);
sumn = sumn + term;
end
y = sumn;
% function [y, k] = exp1(x)
% eˆx ˜ 1 +x +xˆ2/2! +xˆ3/3! + ...
% using nested multiplication
function [y, k] = exp1(x)
sump = 0;
sumn = 1;
k = 0;
term = 1;
while sump ˜= sumn
sump = sumn;
k = k+1;
term = term * x / k;
sumn = sumn + term;
end
y = sumn;
Note the difference between how term is computed in the two functions. The func-
tion to the left (exp0) requires exponention and a call to factorial, while the one
to the right (exp1), requires only multiplication instead.
Tut2 – Computer arithmetic 21 c© C. Christara, 2012-22
% function [y, k] = exp2(x)
% eˆx ˜ 1 +x +xˆ2/2! +xˆ3/3! + ...
% using nested mult. if x >= 0,
% and eˆx ˜ 1/eˆ{-x} if x < 0
function [y, k] = exp2(x)
sump = 0;
sumn = 1;
k = 0;
term = 1;
ax = abs(x);
while sump ˜= sumn
sump = sumn;
k = k+1;
term = term * ax / k;
sumn = sumn + term;
end
if (x >= 0) y = sumn;
else y = 1/sumn; end
% script for testing various
% approximations to eˆx, x = -22:22
k = 0;
for x = -22:22
k = k + 1;
err0(k) = (exp(x)-exp0(x))/exp(x);
err1(k) = (exp(x)-exp1(x))/exp(x);
err2(k) = (exp(x)-exp2(x))/exp(x);
fprintf(’%4d %10.2e %10.2e %10.2e\n’,
x, err0(k), err1(k), err2(k));
end
The function to the left (exp2) is the same as
exp1, except that exp2 works on |x| instead of
x, and the case x < 0 is handled by 1/e|x|.
Tut2 – Computer arithmetic 22 c© C. Christara, 2012-22
-22 1.2e+02 8.1e+00 1.9e-16
-21 -3.5e+01 5.2e+00 -2.7e-16
-20 -1.0e+00 -1.7e+00 -2.0e-16
-19 5.4e-01 -5.8e-01 1.5e-16
-18 -4.9e-02 -1.0e-02 2.2e-16
-17 -1.0e-03 4.1e-03 8.0e-16
-16 -2.9e-04 -5.9e-05 0.0e+00
-15 -1.0e-05 -2.3e-05 3.5e-16
-14 8.6e-06 -4.3e-07 3.8e-16
-13 1.3e-06 -1.6e-06 0.0e+00
-12 -6.1e-08 -1.6e-07 -4.1e-16
-11 -7.6e-08 -2.8e-08 0.0e+00
-10 7.2e-09 3.1e-09 -3.0e-16
-9 5.5e-10 1.4e-10 2.2e-16
-8 1.5e-10 6.6e-12 -1.6e-16
-7 -1.3e-11 1.8e-11 -8.3e-16
-6 7.3e-13 -7.1e-13 0.0e+00
-5 -2.1e-13 2.1e-13 -3.9e-16
-4 -1.4e-14 -1.7e-14 3.8e-16
-3 -8.4e-16 -1.7e-15 -2.8e-16
-2 -4.1e-16 -4.1e-16 -2.1e-16
-1 -3.0e-16 -3.0e-16 1.5e-16
0 0.0e+00 0.0e+00 0.0e+00
1 0.0e+00 0.0e+00 0.0e+00
2 2.4e-16 2.4e-16 2.4e-16
3 3.5e-16 3.5e-16 3.5e-16
4 -5.2e-16 -5.2e-16 -5.2e-16
5 1.9e-16 3.8e-16 3.8e-16
6 0.0e+00 0.0e+00 0.0e+00
7 6.2e-16 8.3e-16 8.3e-16
8 1.5e-16 1.5e-16 1.5e-16
9 0.0e+00 0.0e+00 0.0e+00
10 3.3e-16 3.3e-16 3.3e-16
11 0.0e+00 0.0e+00 0.0e+00
12 3.6e-16 3.6e-16 3.6e-16
13 1.3e-16 0.0e+00 0.0e+00
14 -1.9e-16 -3.9e-16 -3.9e-16
15 0.0e+00 -2.8e-16 -2.8e-16
16 0.0e+00 0.0e+00 0.0e+00
17 -3.1e-16 -7.7e-16 -7.7e-16
18 0.0e+00 -2.3e-16 -2.3e-16
19 1.7e-16 0.0e+00 0.0e+00
20 2.5e-16 1.2e-16 1.2e-16
21 3.6e-16 1.8e-16 1.8e-16
22 -2.7e-16 -2.7e-16 -2.7e-16
Tut2 – Computer arithmetic 23 c© C. Christara, 2012-22
Notes:
Between exp0 and exp1, we notice that there are cases where the error of exp0
is larger than the error of exp1, and vice-versa. However, among the first cases, we
see some in which there is noticeable difference (more than one order) in the errors,
while in the second cases, the difference in the errors is small (half an order or less).
Avoiding the more complicated operations (exponentiation and factorial) reduces the
probability of large errors.
Between exp1 and exp2, we notice that, for quite negative x, exp1 exhibits huge
errors, while exp2 matches MATLAB’s exp to machine precision (relative error to
the level of machine epsilon). Note that exp1 uses alternating signs in the summation,
which inhibit the danger of catastrophic cancellation, while exp2 avoids this problem.
Aside 1: Saturation can, sometimes, cause large errors. For example, in the operation
1 + x, the x will be lost if x < ǫmach, and the 1 will be lost if x > 1/ǫmach. We need
to exercise caution in such operations. However, in the question of approximating ex
by Taylor series, saturation serves as a convenient technique to let us (and the code)
know where to stop the summation.
Aside 2: We would not be able to do the summation from the smallest to the largest
term, because we do not know which is the smallest term needed.
案例分类
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