Explain propagation of errors in the diffedence of two quantities and also in the divison of two quantities.

Explanation

Error in the differnce of two quantities:
Let `DeltaA` and `DeltaB` be the absolute errors in the two quantities A and B respectgively. Then,
Measured value of `A=A+-DeltaA`
Measured value of `B=B+-DeltaB`
Consider the difference `Z=A-B`
The error `DeltaZ` in Z is then given by
`Z+-DeltaZ=(A+-DeltaA)-(B+-DeltaB)`
`=(A-B)+-(DeltaA+DeltaB)`
`=Z+-(DeltaA+DeltaB)`
(or) `DeltaZ=DeltaA+DeltaB`
The maximum error in difference of two quantities is equal to the sum of the absolute errors in the individual quantities.
Error in the division or quotient of two quantities.
Let `DeltaA` and `DeltaB` be the absolute erros in the two quantities A and B respectively.
Consider the quotient `Z=A/B`
The error `DeltaZ` in Z is given by
`Z+-Z=(A+-DeltA)/(B+-DeltaB)=(A(1+-(DeltaA)/A))/(B(1+1(DeltaB)/B))`
`=A/B(1+-(DeltaA)/A)(1+-(Delta+-(DeltaB)/B)^(-1)`
or `Z+-DeltaZ=Z(1+-(DeltaA)/A)(1+-(DeltaB)/B)`
[using `(1+x)^(n)~~1+nx`, when `x lt lt 1`]
Dividing both sides by Z, we get
`1+-(DeltaZ)/Z=(1+-(DeltaA)/A)(1+-(DeltaB)/B)`
`=1+-(DeltaA)/A+-(DeltaB)/B+-(DeltaA)/A . (DeltaB)/B`
As the terms `DeltaA//A` and `DeltaB//B` are small, their product term can be neglected.
The maximum fractional error in Z is given by
`(DeltaZ)/Z=((DeltaA)/A+(DeltaB)/B)`
The maximum fractional error in the quotient of two quantities is equal to the sum of their individual fractional errors.