Errors
Errors are unavoidable due to the nature of measurements. It shows how reliable the data that was obtained can be. Hence, it is important to know the error of the physical quantity that we are dealing with. However, most of the times, the error of the physical quantity measured is not directly obtained from the measurements but rather calculated from each individual components of it. In this post, we aim to show you the two most common method to obtain the error of a physical quantity: The Maximum-Minimum method and the Derivative methodMethods
Before we start deriving the errors, we need to establish a consensus. Suppose we have two variables a and b with their uncertainty being ∆a and ∆b and two random constants being x and y. For the Derivative method specifically, we assume t is a dummy variable (a varibale having no significance but for mathematical rigor) and da⁄dt = ∆a and db⁄dt = ∆b. Now, we can start deriving the errors.The Derivative method
The definition of a derivative is, by definition: limΔx→0 f(x+Δx)-f(x)⁄Δx = df(x)⁄dx, in where the result is obtained by ignoring terms of Δx with powers no less than 2 due to their non-existent effect on the result. In a similar fashion, instead of an infinitesimal presented here, a finitely small quantity (Δx >> x) takes the place and the argument is the same with that of infinitesimals.Common functions
a±b
We have: d(a±b)⁄dt = da⁄dt ± db⁄dt = ∆a ± ∆b. Since we aim to maximise the error, the error is, effectively ∆a + ∆b.ab
We have: d(ab)⁄dt = adb⁄dt + bda⁄dt = a∆b + b∆a. Hence, the error is a∆b + b∆aa⁄b
We have: d(a⁄b)⁄dt = (bda⁄dt - adb⁄dt)⁄b2= (b∆a - a∆b)⁄b2. Since we aim to maximise the error, the error is, effectively (b∆a - a∆b)⁄b2logxa
We have: d(logxa)⁄dt = d(lna)⁄lnx × dt = 1⁄lnx × a × da⁄dt = ∆a⁄lnx × a. Hence, the error is ∆a⁄lnx × a.ax
We have d(ax)⁄dt= xa(x-1) da⁄dt = xax-1 ∆a. Hence, the error is xax-1 ∆axa
We have xa = ealnx. Therefore d(xa)⁄dt = d(ealnx)⁄dt = d(alnx)⁄dt× ealnx = lnx×da⁄dtealnx = xalnx ∆a. Hence, the error is: xalnx ∆alogxa
We have logxa = lna⁄lnx. Therefore, d(logxa)⁄dt = dlna⁄lnx × dt = da⁄a × lnx × dt = ∆a⁄lnx × a. Hence, the error is: ∆a⁄lnx × a.f(a)
Suppose f(a) is differentiable, we have: df(a)⁄dt = df(a)⁄da × da⁄dt = f'(a) × ∆a. Hence, the error is: f'(a) × ∆a.Generalised functions
axby
We have: d(axby)⁄dt = d(ax)⁄dtby + axd(by)⁄dt = xa(x-1)by da⁄dt + ax (yby-1) db⁄dt = xa(x-1)by ∆a +yaxb(y-1) ∆b. Hence, the error is: xa(x-1)by ∆a +yaxb(y-1) ∆b.ab
We have: ab = ebln(a). Therefore: d(ab)⁄dt = d(eblna)⁄dt = ebln(a) d(blna)⁄dt = ab (b×da⁄dt × a + ln(a)db⁄dt) = ab(b + ln(a)∆b). Hence the error is ab(b∆a⁄a + ln(a)∆b).logba
We have: logba = lna⁄lnb. Therefore, d(logba)⁄dt = d(lna⁄lnb)⁄dt = (da⁄dt lnb - lna db⁄dt)⁄ln2b = (∆a lnb - ∆b lna)⁄ln2b. Since we aim to maximise the error, the error is effectively: (∆a lnb + ∆b lna)⁄ln2bf(a,b)
We have: df⁄dt = ∂f⁄∂a×da⁄dt + ∂f⁄∂b×db⁄dt= ∂f⁄∂a×∆a + ∂f⁄∂b×∆b. Hence, the error is ∂f⁄∂a×∆a + ∂f⁄∂b×∆bThe Maximum-Minimum method
The essence behind this method is to examine the discrepancies between the true value and the maximum and minimum possible measured value. The error of the quantity is the maximum of the two discrepancies. It is simple in principle but difficult in practice. Lets examine some functions.Common functions
a±b
The maximum possibly measured value of it is (a + Δa) + (b + Δb) = (a + b) + (Δa + Δb).Hence, the discrepancy is: | (a + b) + (Δa + Δb) - (a + b) | = (Δa + Δb).
The minimum possibly measured value of it is (a - Δa) - (b + Δb) = (a + b) - (Δa + Δb).
Hence, the discrepancy is: | (a + b) - (Δa + Δb) - (a + b) | = (Δa + Δb).
Therefore, the error for this function is (Δa + Δb).
ab
Without loss of generality, we divide the quantity into two cases: Either- a and b have different signs
- a and b have the same sign
Then, the maximum possibly measured value of it is (a + Δa)(b + Δb) = ab + (bΔa + aΔb) + ΔaΔb.
Hence, the discrepancy is: | ab + (bΔa + aΔb) + ΔaΔb - ab | = (bΔa + aΔb) + ΔaΔb.
The minimum possibly measured value of it is (a - Δa)(b + Δb) = ab - (bΔa + aΔb) + ΔaΔb.
Hence, the discrepancy is: | ab - (bΔa + aΔb) + ΔaΔb - ab | = -(bΔa + aΔb) + ΔaΔb.
Therefore, the error for this function is (bΔa + aΔb) + ΔaΔb. For the second case, suppose a is the negative quantity.