Propagation of Error with Single and Multiple Independent Variables

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1 Propagation of Error with Single an Multiple Inepenent Variables Jack Merrin February 11, Summary Often it is necessary to calculate the uncertainty of erive quantities. This proceure an convention is calle propagation of error. Calculating the propagation of error formula requires knowing how to take erivatives. We give the propagation of error formulas for a single variable an multiple inepenent variable functions. We list some common error propagation formulas that work in most cases. Step by step error propagation can often be use as a shortcut to calculate the uncertainty. If we know the error propagation formula we can fin the terms that are causing the most uncertainty an try to reuce these ominant errors. It is useful to write a short MATLAB script to calculate error propagation, especially when the formula is complex or a lot of significant igits are involve. How error propagates As a math problem, I ask you to a 5. an The result is straightforwar The numbers have ifferent numbers of igits an precision. 5. is precise to the tenths while is precise to the hunreths. This is not a problem in mathematics, but the situation is more subtle in error analysis. To see what is going on you nee to look at the uncertainties. Now, if I ask you to a 5.±1. an 10.11±0.76 then what o you o? The first thing you want to ask yourself is if these numbers are properly reporte measurements. After a quick check, we see there are two igits of uncertainty that match the precision of the measurement. In error analysis, you still a the numbers to fin the sum, but to fin the combine error we nee a propagation of error formula. In this situation, the rule is σ A+B = σa + σ B = This chapter is all about fining the rules like this. We can now calculate the uncertainty σ A+B = an roun to two significant igits. Then 1

2 we match the precision of the calculate quantity to the uncertainty. Our final result is that (5. ± 1.) + (10.11 ± 0.76) = ± ± 1.4 Notice that if you always start with uncertainties that have two significant igits, you get out a result that has two significant igits of uncertainty. You on t have to worry that the first number 5. has only two significant igits, but the final result 15.3 has three significant igits. This comes about naturally by the golen rule of two significant igits of uncertainty. There are error propagation rules for single variable functions an multiple variable functions. The single variable situation is not too complicate. You just nee to unerstan ifferential calculus. For the situation with multiple variables functions, you have to worry if the variables are correlate or inepenent. In most cases, we on t have to worry about correlations an the formulas are straightforwar. 3 Rules of ifferential calculus refresher We now o a little review of ifferentiation because we nee erivatives for error propagation. Orinary erivatives are neee for a single variable function, an partial erivatives are neee for multiple variable functions. The erivative is efine in calculus as f (x) = f x = lim f(x + x) f(x) x 0 x The following list contains the basic formulas to master ifferential calculus. c = 0 x = 1 (x n ) = nx n 1 x = x x (cu) = cu (c 1 u + c v) = c 1 u + c v (uv) = u v + uv (1/u) = u /u (u/v) = u v uv f(u(x)) = f u ux = f (u)u (x) (e x ) = e x (a x ) = a x ln a (ln x) = 1 x sin x = + cos x x cos x = sin x x sinh x = + cosh x x x tan x = + sec x x cot x = csc x x tanh x = +sech x v (log a x) = 1 x ln a sec x = + sec x tan x x csc x = csc x cot x x sech x = sech x tanh x x

3 cosh x = + sinh x x x arcsin x = x x arccos x = 1 1 x x arsinh x = + 1 x + 1 x arcosh x = + 1 x 1 x coth x = csch x csch x = csch x coth x x x arctan x = x x arcsec x = + 1 x x 1 x arccot x = x x arccsc x = 1 x x 1 x artanh x = x x arsech x = 1 x 1 x x arcoth x = x x arcsch x = 1 x 1 + x The partial erivative is like the normal erivative except you hol the other variables constant when you take it. Here is the efinition (x, y) x (x, y) y f(x + x, y) f(x, y) = lim x 0 x f(x, y + y) f(x, y) = lim y 0 y For example, if we wante to fin all the partial erivatives of f(x, y, z) = xy z 3 we woul use the power rule (x n ) = nx n 1. There are three possible partial erivatives x = y z 3 y = xyz3 z = 3xy z 4 Error propagation with one variable If we have measure Q ± σ Q then the uncertainty in Q(A) is given by σ Q = Q (A) σ A I assume you are familiar with ifferentiation. One can use a program like Mathematica or your graphing CAS calculator to fin erivatives. Example 4.1. The volume of a cube from error propagation. Suppose a machinist has constructe a precise cube on a milling machine, an you want to fin its volume. You repeately measure one of the sies to be s = ± cm What is the volume an its uncertainty assuming the length, with, an height are ientical? 3

4 Solution 4.1. We calculate the volume as V = s 3 an wait to roun until we know the uncertainty. V = (1.053 cm) 3 = cm 3 To fin the uncertainty we use the error propagation formula σ V = V (s) σ s = 3s σ s σ V = 3(1.053 cm) (0.010 cm) = cm 3 After rouning we get V = 1.168(34) cm 3. 5 Error propagation with multiple inepenent variables Often we want to calculate some quantity that requires a series of iniviual measurements that must be combine. For example, if we wante to calculate A/B we woul first measure A then measure B an then we coul calculate A/B. If we are oing error analysis, then to fin the error in A/B we will nee to apply a formula of error propagation. There is a significant simplification if the two quantities are thought to be inepenent like the charge an mass of an electron. The error propagation formula for multiple inepenent variables labele a 1, a,..., a M is given by ( ) M σf = σa a j j j=1 We will often use the notation that a 1 = A a = B for simplification when we o calculations. Propagation of error for two inepenent variables is given by ( ) ( ) σf = σa + σb A B Example 5.1. Show that σ A±B = σ A + σ B Solution 5.1. Evaluating the partial erivatives (A ± B) A = 1 (A ± B) B = ±1 Applying the error propagation formula we have σ A±B = (1) σ A + (±1) σ B = σa + σ B 4

5 Example 5.. Fin the perimeter of a rectangle of l = 1.5() an w = 4.44(33) Solution 5.. The perimeter is equal to P = (l + w) We can fin the perimeter by plugging in the values P = ( ) = We can fin the uncertainty by applying the propagation of error formula σ P = σ P = ( ) P σl + l ( ) P σw w 4σ l + 4σ w = = Our final answer is P = ± 0.79 σ P 0.79 Example 5.3. If Q = AB, then show that σa σ AB = AB A + σ B B Solution 5.3. Evaluating the partial erivatives (AB) A = B (AB) B = A Applying the error propagation formula we have σ AB = B σa + A σb Divie both sies by AB. σ AB AB = 1 B AB σa + A σb = σa A + σ B B Example 5.4. If Q = A/B, then show that σa σ AB = A/B A + σ B B Solution 5.4. Evaluating the partial erivatives (A/B) A = 1/B (A/B) B = A/B 5

6 Applying the error propagation formula we have σ A/B = (1/B) σa + (A /B 4 )σb Divie both sies by A/B. σ A/B A/B = 1 (1/B) A/B σa + (A /B 4 )σb = σa A + σ B B 6 Common error formulas Q = A + B then σ Q = σ A + σ B Q = A B then σ Q = σa + σ B Q = AB then σ Q σ Q = A A + σ B B Q = A/B then σ Q σ Q = A A + σ B B Q = AB/C... then σ Q Q = σ A A + σ B B + σ C C +... Q = A a B B /C c... then σ Q Q = a σa A + b σb B + c σc C +... Many formulas in physics are of the last form. For example, F = GmM/r or ω = k/m. 7 Error propagation step by step Suppose we want to compute the error in the formula Q(A, B, C, D) = A + B C + D 6

7 It is convenient to o it step by step. Fin the uncertainty in the numerator an enominator iniviually then combine the results. We can let Y = A + B an Z = C + D then fin the uncertainties in Y an Z first. σ Y = σ A + σ B σ Z = σ C + σ D We can then calculate the uncertainty for Y/Z which woul be σ Q Q = σy Y + σ Z Z You can check for yourself that you get the same result as if you i the calculation in one step. If the formula was f = A B A + B then we couln t o it step by step the same way because the numerator an the enominator contain the same variables that repeat. The partial erivatives of the numerator an enominator cannot be separate. 8 The ominant error The error propagation formula tells you where you shoul focus your efforts if you want to reuce the uncertainty in a erive quantity. Example 8.1. Suppose you are calculating g from experiments with a penulum with small oscillations. ω = g/l g = 4π l T If the fractional uncertainty in the length an perio are both ten percent, then how shoul you procee to improve the experimental etermination of g? Solution 8.1. The error propagation formula for this situation. easily write own the error propagation formula σ g g = σl l + 4σ T T We can If the fractional uncertainty in the length is 10 percent an the fractional uncertainty in the perio is 10 percent then the expression in the square root is Clearly measuring the perio more precisely will have the largest effect on reucing the ominant error in the uncertainty. 7

8 9 MATLAB examples Example 9.1. Fining the uncertainty in e N. Q(N) = e N an the uncertainty. Let N = 3.54(35) fin Solution 9.1. We know that (e x ) = e x. So we easily calculate the uncertainty using σ Q = e N σ N = e N σ N Here is what the Matlab coe looks like to calculate that. format long; N = 3.54; sign = ; f = exp(n) sigf = exp(n)*sign >> f = >> sigf = We o the rouning an fin f = 5.85(90) Example 9.. Fining the uncertainty in sin(θ). What is the uncertainty in Q = sin(θ) when θ = 30.0 ±.5? Solution 9.. First we convert the numbers to raians. The error propagation formula is then σ Q = cos θ σ θ It is easy to o all these calculations in Matlab. format long; t = 30; sigt =.5; tra = 30 * pi/180; sigtra =.5 * pi/180; f = cos(tra)*sigtra f = sin(tra) >> >> So the answer is sin θ = 0.500(38). Example 9.3. Fining the Stefan Boltzmann constant an its uncertainty. What is the value an uncertainty of the Stefan Boltzmann constant σ = π k 4 B 60 h 3 c 8

9 Solution 9.3. It s funny that the notation for the Stefan Boltzmann constant is alreay σ, on t get confuse. The propagation of error formula is a common one. σ σ = σ 4 σ k B k B + 3 σ h h Since c is efine it has no uncertainty. Also constant factors like π or have no uncertainty. A quick google for NIST hbar, NIST c, an NIST Boltzmann s constant gives the necessary ata. format long; hb = e-34; shb = e-34; c = ; kb = e-3; skb = e-3; SB = pi^/60 * kb^4/(c^*hb^3) sigsb = SB*(16*skb^/kb^+9*shb^/hb^)^(0.5) >> e-08 >> e-13 The uncertainty is J m s 1 K 4. So the final result is σ = (13) 10 8 Jm s 1 K 4 This calculation agrees with the NIST 016 value. 9

Calculus Calculating the Derivative Chapter 4 Section 1 Techniques for Finding Derivatives

Calculus Calculating the Derivative Chapter 4 Section 1 Techniques for Fining Derivatives Essential Question: How is the erivative etermine of a single term? Stuent Objectives: The stuent will etermine