Question

1/z =1/a+1/b find relative error in z

Answer 1

Answer:

solved

Explanation:

1/z =1/a+1/b = a+b/ab

z = ab/a+b

logz = loga + logb - log a+ b

differentiate both sides

Δz/z = Δa/a + Δb/b - Δ(a+b) / (a+b)

Δz/z = Δa/a + Δb/b - (Δa+Δb) / (a+b)

Answer 2

Explanation:

it is given that, 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

it means, relative error of z = relative error of A + relative error of B

hence proved

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