Q. What is the approximate error in f(x) if x = 1000 ± 10% ?
a. zero b. ± 5% c. ± 10% d. ± 15% e. ± 20%A. The correct answer is:
Observe that the value of f(x) will be close to 1 for any large value of x. Small variations in x should not significantly affect this result.
"Wait!", stammered the startled student (remembering an earlier lesson), "When dividing two values, the rule is to add the percentage errors, and that gives ± 20%".
Who's right? What's wrong?
METHOD 1 Compute the derivative df(x)/dx
Using the calculus, df(x) = (x+1)-2 dx
For x = 1000, f(x) = 0.9990 ± 0.0001 This supports the near-zero error result.
METHOD 2 Try the worst case situations.
The worst case error, for x=900 or x=1100, gives
f(1100) = 1100/1101 = 0.999 f(1000) = 1000/1001 = 0.999 f( 900) = 900/901 = 0.999Which shows that variations in x really do not have much effect on the result.
METHOD 3 Use the shortcut rules for add, subtract, multiply, divide.
"When dividing two values, add the percentage errors in each"
Since the numerator and the denominator both have a 10% error, the total error should be 10 + 10 = 20%.
THIS IS CLEARLY INCORRECT, BUT WHY?
This shortcut method assumes that the numerator and denominator are unrelated values. This method works as expected for functions such as g(P,Q) = P/(1+Q). It fails in the case of f(x) = x/(1+x) because the same value with the same error is found twice.
To get the worst error in g(P,Q), P must be high when Q is low, or vice-versa.
g(1100, 900) = 1100/901 = 1.221 g(1000,1000) = 1000/1001 = 0.999 g( 900,1100) = 900/1101 = 0.817Which shows that the 20% error estimate really was correct under these conditions.
This situation can never happen for f(x). The appearance of x in both the numerator and the denominator ensures that any small error cancels out to give a negligible error.
Note that even in this example with f(P,Q), it is assumed that there is no hidden relationship connecting P and Q. If it was the case, for example, that P = 2Q, the results would be different.
Similarly, the derivative method of finding errors relies on an underlying assumption that different variables are uncorrelated.
Mathematics is a tool and, like laboratory instruments, knowing your tools is key to using them correctly.
Consider the case of: f(t) = t + t.
This may also be written: f(t) =2 t
If t has an error of 10%, then 2t has an error of 10%,
the error in t+t is not 20%.
Parallel resistors present a challenge for error analysis because the value of each resistor appears in both the numerator and the denominator.
This is a case where applying the simple rules for adding, multiplying, and dividing errored terms does not work.
As an example, let A and B be 1000 ohm resistors, each with a 10% tolerance.
Rtotal = 500 ohm
What is the error in Rtotal ?
Using a worst case analysis:
R(1100,1100) = (1100 x 1100)/(1100 + 1100) = 550 R(1100, 900) = (1100 x 900)/(1100 + 900) = 495 R(1000,1000) = (1000 x 1000)/(1000 + 1000) = 500 R( 900,1100) = ( 900 x 1100)/( 900 + 1100) = 495 R( 900, 900) = ( 900 x 900)/( 900 + 900) = 450
These cases show 450 < Rtotal < 550, for about 10% error.
Parallel 10% tolerance resistors simply give 10% overall error.
A full solution requires computing the derivatives and adding the error contributions from resistors A and B.
This gives error = ( A2 dB + B2 dA ) / (A + B)2
In this example, A = 1000 ± 100, and B = 1000 ± 100,
error = ( 10002 x 100 + 10002 x 100 ) / (1000 + 1000)2 = 50
The result is Rtotal = 500 ± 50 ohm, as expected.