f a f a X Y f( )a ( ) ( ) af fay y= = Figure 2.2: Errors in a pratical situation f

Lecture 3 (Tue, Aug. 28). Conditioning
Consider a general problem (f(x)) which is well defined. Suppose we have the input x = a
and we need to find the result (answer) y = f(a). Now suppose we use a method ˆ f(x) to
approximate f(x) with an input â that is close to a with an absolute backward error |a − â|.
Here |·|| is a certain measure (norm). Then with a certain measure (norm) |·|| that might be
different from the one for a, the absolute error of the result is |f(a) − ˆ f(â)|, which is called
the absolute forward error.
The error in the result (by ignoring the norm) can be decomposed into two parts (Fig-
ure 2.2):
f(a) − ˆ f(â) = f(a) − f(â)
+ f(â) −
Propagated data error
ˆ f(â)
Computational error
The computational error is caused by the approximation of the method. The propagated
Figure 2.2: Errors in a pratical situation
a
a
X
f
f
f
Y
y = fa ( )
f( a)
y= f(
a)
data error is caused the sensitivity of the problem itself.
The conditioning of a problem f is about to sensitivity of the result f(a) to a small
change in a. The absolute condition number κf(a) is defined by
κf(a) = lim sup â→a
If f is a function and is differentiable at a, then
κf(a) = |J(a)|,
|f(a) − f(â)|
.
|a − â|
1

2
where J(x) is the Jacobian of f(x). Obviously, for any â close to a,
|f(a) − f(â)| ≈ κf(a)|a − â|.
So κf(a) is large, a small change in a may result in a large change in f(a). In this case, we
say the problem is ill-conditioned. Otherwise it is well-conditioned.
Similarly, when a = 0 and f(a) = 0, using the relative backward error |a − â|/|a| and
relative forward error |f(a) − f(â)|/|f(a)|, we can define the relative condition number
˜κf(a) = lim sup â→a
If f(x) is differentiable at a, then
When â is close to a,
|f(a) − f(â)|/|f(a)|
|a − â|/|a|
˜κf(a) = |a|
|f(a)| |J(a)|.
|f(a) − f(â)|
|f(a)|
≈ ˜κf(a)
|a − â|
.
|a|
= |a|
|f(a)| κf(a).
Again, ˜κf(a) measures the sensitivity of f(x) at a, but in a relative sense.
Example 3.1 Let f(x) = √ x. Determine κf and ˜κf at a > 0.
Solution. Since J(a) = f ′ (x) = 1/(2 √ a),
κf(a) = |J(a)| = 1
2 √ a , ˜κf(a) = a
√ |J(a)| =
a a
√
a
1
2 √ a
= 1
2 .
Hence, in absolute sense, the problem is ill-conditioned when a is close to 0, and it is well
conditioned when a is away from 0, say a ≥ 1.
In relative sense, the problem is well-conditioned for any a > 0 (even for a = 0 in the
limit sense).
Example 3.2 In this example we investigate how a root of a polynomial will change when
an error is introduced to a particular coefficient. Let p(x) = a0 + a1x + . . . + anx n and λ be
a single root of p(x) = 0. (That means p(x) = (x − λ)q(x) and q(λ) = 0.) It is well-known
that λ is a function of the coefficients a0, . . . , an. In particular, λ is a function of ai for some
i when other coefficients remain unchanged. Let λ = f(ai). Determine κf(ai) and ˜κf(ai).
Solution. The function λ = f(ai) is a single variable function. Since λ is a single root of
p(x), f(ai) is differentiable. We need to find f ′ (ai).
Because λ is a root of p(x) = 0,
p(λ) = a0 + a1λ + . . . + ai−1λ i−1 + aiλ i + ai+1λ i+1 + . . . + anλ n = 0.
Differentiating both sides with respect to ai we have
[a1 + . . . (i − 1)ai−1λ i−2 + iaiλ i−1 + (i + 1)ai+1λ i + . . . + nanλ n−1 ]f ′ (ai) + λ i = 0,

or equivalently
So
and
p ′ (λ)f ′ (ai) = −λ i ⇒ f ′ (ai) = − λi
p ′ (λ) .
κf = |f ′ (ai)| = |λ|i
|p ′ (λ)| ,
˜κf = κ |ai|
|λ|
i−1 |ai||λ|
=
|p ′ .
(λ)|
Roughly speaking, the problem of finding a root of a polynomial equation is ill-conditioned
if p ′ (λ) is small. In particular, when λ is a multiple root, p ′ (λ) = 0 and κf = ∞ (˜κf = ∞).
3