1. Sec:4.2
ERROR PROPAGATION

2. Sec:4.2 ERROR PROPAGATION
The purpose of this section is to study how errors in numbers can propagate through mathematical functions.
Why?
Example: =
round-off errors are directly related to the manner in which numbers are stored in a computer.
* Numbers such as π, e, or cannot be expressed by a fixed number of significant figures
*numbers on the computer are represented with a binary, or base-2, system.
*binary system 
floating point numbers ⊂𝑅𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
*the gap between adjacent numbers are in a relative sense never larger that eps (machine epsilon). Ex: (1+eps/2)-1
Error =
relative error = =2.1𝑒−16
*round-off errors originate from the fact that computers retain only a fixed number of significant figures during a calculation.

3. ∆𝒇 𝒙 =𝒇′( 𝒙 )𝚫 𝒙 Sec:4.2 ERROR PROPAGATION
Assume that 𝒙 is an approximation of x.
Assume that 𝒙 is an approximation of x.
∆ 𝒙 =𝒙− 𝒙
∆ 𝒙 =𝒙− 𝒙
Suppose that we want to calculate 𝑓(𝒙)
Suppose that we want to calculate 𝟐𝒙
we would like to estimate
𝒇(𝒙)−𝒇( 𝒙 )
𝑳𝒆𝒕: ∆𝒇 𝒙 =𝒆𝒔𝒕𝒊𝒎𝒂𝒕𝒆 𝒐𝒇 𝒇(𝒙)−𝒇( 𝒙 )
𝟐∆ 𝒙 =𝟐𝒙−𝟐 𝒙
Zero-order approximation of 𝒇
the error of the multiplication by 2 is double the error in x
𝒇 𝒙 =𝒇 𝒙 +𝒇′(𝝃)(x- 𝒙 )
Rearranging terms yields
𝒇 𝒙 − 𝒇 𝒙 =𝒇′(𝝃)(x- 𝒙 )
Example:
Given a value of 𝑥 = 2.5 with an error of ∆ 𝑥 = 0.01, estimate the resulting error in the function, f (x) = 𝑥 3 .
=𝒇′(𝝃)𝚫 𝒙
≈𝒇′( 𝒙 )𝚫 𝒙
Error Propagation in a Function of a Single Variable
∆𝒇 𝒙 =𝒇′( 𝒙 )𝚫 𝒙
∆𝒇 𝒙 =𝒇′( 𝒙 )𝚫 𝒙
=𝟑 (2.5) 𝟐 𝟎.𝟎𝟏 =𝟎.𝟏𝟖𝟕𝟓

4. Sec:4.2 ERROR PROPAGATION
If we have a function of two independent variables u and v, the Taylor series can be written as
𝑓 𝑢 𝑖 +ℎ, 𝑣 𝑗 +𝑘 =𝑓 𝑢 𝑖 , 𝑣 𝑗 + 𝜕𝑓 𝜕𝑢 ℎ+ 𝜕𝑓 𝜕𝑣 𝑘+ 1 2! 𝜕 2 𝑓 𝜕 𝑢 2 ℎ 2 + 𝜕 2 𝑓 𝜕 𝑣 2 𝑘 2 + 𝜕 2 𝑓 𝜕𝑢𝜕𝑣 ℎ𝑘 +…
𝑓 𝑢 𝑖+1 , 𝑣 𝑗+1 =𝑓 𝑢 𝑖 , 𝑣 𝑗 + 𝜕𝑓 𝜕𝑢 𝑢 𝑖+1 − 𝑢 𝑖 + 𝜕𝑓 𝜕𝑣 (𝑣 𝑗+1 − 𝑣 𝑗 )
+ 1 2! 𝜕 2 𝑓 𝜕 𝑢 𝑢 𝑖+1 − 𝑢 𝑖 𝜕 2 𝑓 𝜕 𝑣 2 (𝑣 𝑗+1 − 𝑣 𝑗 ) 2 + 𝜕 2 𝑓 𝜕𝑢𝜕𝑣 𝑢 𝑖+1 − 𝑢 𝑖 (𝑣 𝑗+1 − 𝑣 𝑗 ) +…
∆𝒇 𝒖, 𝒗 = 𝝏𝒇 𝝏𝒖 𝚫 𝒖 + 𝝏𝒇 𝝏𝒗 𝚫 𝒗
∆𝒇 𝒙 𝟏 , 𝒙 𝟐 ,…, 𝒙 𝒏 = 𝝏𝒇 𝝏 𝒙 𝟏 𝚫 𝒙 𝟏 + 𝝏𝒇 𝝏 𝒙 𝟐 𝚫 𝒙 𝟐 +…+ 𝝏𝒇 𝝏 𝒙 𝒏 𝚫 𝒙 𝒏

5. Sec:4.2 ERROR PROPAGATION
∆𝒇 𝒙 𝟏 , 𝒙 𝟐 ,…, 𝒙 𝒏 ≈ 𝝏𝒇 𝝏 𝒙 𝟏 𝚫 𝒙 𝟏 + 𝝏𝒇 𝝏 𝒙 𝟐 𝚫 𝒙 𝟐 +…+ 𝝏𝒇 𝝏 𝒙 𝒏 𝚫 𝒙 𝒏
Example:
𝑭(𝒙,𝒚,𝒛)= 𝒙 𝒚 𝟑 𝟖𝒛
Given:
𝑥 = with ∆ 𝑥 = 30,
𝑦 = with ∆ 𝑦 = 0.03,
𝑧 = with ∆ 𝑧 = ,
estimate the resulting error in the function, F

6. ≈ Sec:4.2 ERROR PROPAGATION Example:
Stability and Condition
The condition of a mathematical problem relates to its sensitivity to changes in its input values.
Example:
How small is
𝜋 𝑥 𝑥−22=0
𝑥 1
𝑥 2
𝑥 1 − 𝑥 1
𝑥 1
( …3) 𝑥 𝑥−22=0
𝑥 2
𝑥 2 − 𝑥 2
We would like to drive an equation similar to :
We say that a computation is numerically unstable if small changes in the input values produce correspondingly large changes in the final results
Condition Number:
Relative error in 𝑓(𝑥)
constant
Relative error in 𝑥 ≈

7. ≈ Sec:4.2 ERROR PROPAGATION 𝒙 𝒇′( 𝒙 ) 𝒇 𝒙
We would like to drive an equation similar to :
Condition Number:
Condition Number:
𝒙 𝒇′( 𝒙 ) 𝒇 𝒙
Relative error in 𝑓(𝑥)
constant
Relative error in 𝑥 ≈
The condition number provides a measure of the extent to which a change in x is magnified by f (x).
A value of 1 tells us that the function’s relative error is identical to the relative error in x.
A value greater than 1 tells us that the relative error is amplified
Zero-order approximation of 𝒇
𝒇 𝒙 =𝒇 𝒙 +𝒇′(𝝃)(x- 𝒙 )
𝒇 𝒙 −𝒇 𝒙 𝒇 𝒙 ≈ 𝒙 𝒇′( 𝒙 ) 𝒇 𝒙 𝒙− 𝒙 𝒙
Functions with very large values are said to be ill-conditioned.

8. ≈ ≈ Sec:4.2 ERROR PROPAGATION Problem 4.12/p105: Sol:
Evaluate and interpret the condition numbers for
𝑓 𝑥 = 𝑥−1 +1
𝑥 =
Sol:
𝑓 1 = 1
𝑓 =
𝑥− 𝑥 𝑥 =1𝑒−5
(Relative error in x)
𝑓′ 𝑥 = (𝑥−1)/ 𝑥− 𝑥−1
Cond = 𝑥 𝑓′( 𝑥 ) 𝑓 𝑥 = ∗ =157.6
(condition number)
Relative error in 𝑓(𝑥)
157.6
Relative error in 𝑥 ≈ ≈
(estimate relative error in f(x))
1.576𝑒−3
(actual relative error in f(x))
𝑓 −𝑓 1 𝑓 1 =3.162𝑒−3