Question

Prove that the relative error of a product of three nonzero numbers does not exceed.the sum of the relative errors of the given numbers

Answer 1

Answer:

E/B <  E₂/B₂ + E₁/B₁ + B₃/E₃

Step-by-step explanation:

Let say three numbers  are

A₁ , A₂ , A₃

There closet number B₁ , B₂ , B₃  with Error E₁ , E₂ , E₃  respectively

B₁ * B₂ * B₃ = B

A₁ * A₂ * A₃ = B₁ * B₂ * B₃  + E

A₁ * A₂ * A₃  = (B₁ + E₁)(B₂ + E₂)(B₃ + E₃)

=> A₁ * A₂ * A₃  = (B₁ * B₂ + B₁E₂  + B₂*E₁ + E₁E₂)(B₃ + E₃)

=> A₁ * A₂ * A₃  = B₁ * B₂*B₃ + B₁B₃E₂  + B₂*B₃E₁ + E₁E₂B₃ + B₁ * B₂E₃ + B₁E₂E₃ + B₂*E₁E₃ + E₁E₂E₃

=>  A₁ * A₂ * A₃ - B₁ * B₂*B₃ = B₁B₃E₂  + B₂*B₃E₁ + E₁E₂B₃ + B₁E₂E₃ + B₂*E₁E₃ + E₁E₂E₃

=> E = B₁B₃E₂  + B₂*B₃E₁ + B₁ * B₂E₃ + E₁E₂B₃ + B₁E₂E₃ + B₂*E₁E₃ + E₁E₂E₃

Diving both sides by B₁ * B₂ * B₃ or B

=> E/B = E₂/B₂ + E₁/B₁ + B₃/E₃ + E₁E₂/B₁B₂ + E₂E₃/B₂B₃ + E₁E₃/B₁B₃ + E₁E₂E₃/B₁B₂B₃

x = E₁E₂/B₁B₂ + E₂E₃/B₂B₃ + E₁E₃/B₁B₃ + E₁E₂E₃/B₁B₂B₃

=> E/B = E₂/B₂ + E₁/B₁ + B₃/E₃ + x

x is non zero as numbers are non zero

=> E/B <  E₂/B₂ + E₁/B₁ + B₃/E₃

relative error of a product of numbers < the sum of the relative errors of the given numbers