Question

is the measured values of two quantities are A ± ∆A and B ± ∆A and ∆B being the mean absolute errors. What is the maximum possible error in A±B? show that if z=a/b ∆Z/Z=∆A/A+∆B/B​

Answer 1

if measured values of two quantities are (A ± ∆A) and (B ± ∆B) , where ∆A and ∆B

being the mean absolute error.

we have to find maximum possible error of (A ± B)

for (A ± B)

= (A ± ∆A) ± (B ± ∆B)

= (A ± B) ± (∆A + ∆B)

but for maximum possible error, error of A and B must be added.

so, maximum error of (A ± B) = (∆A + ∆B)

now, z = A/B

taking both sides,

logz = log(A/B)

or, logz = logA - logB

differentiating both sides,

or, dz/z = dA/A - dB/B

but it is assumed that found error must be greater in value. for this in place of negative sign use positive sign.

then, dz/z = dA/A + dB/B

if dz has comparable value of z , dz → ∆z

similarly, dA has comparable value of A, dA → ∆A

dB has comparable value of B, dB → ∆B

so, ∆z/z = ∆A/A + ∆B/B

hence proved