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Math Help - Relative errors product

  1. #1
    Senior Member OReilly's Avatar
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    Relative errors product

    Can someone show me the proof that relative error of product of approximant numbers is equal to sum of relative errors of each factors?

    I need to show that \delta _{x'y'}  = \delta _{x'}  + \delta _{y'}


    I know that
    \left| {xy - x'y'} \right| = \left| {x'} \right|\Delta _{y'}  + \left| {y'} \right|\Delta _{x'}  + \Delta _{x'} \Delta _{y'}

    If we divide both sides with \left| {x'y'} \right| we get that
    \frac{{\left| {xy - x'y'} \right|}}{{\left| {x'y'} \right|}} = \frac{{\left| {x'} \right|\Delta _{y'} }}{{\left| {x'y'} \right|}} + \frac{{\left| {y'} \right|\Delta _{x'} }}{{\left| {x'y'} \right|}} + \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}
    \frac{{\Delta _{x'y'} }}{{\left| {x'y'} \right|}} = \frac{{\Delta _{y'} }}{{\left| {y'} \right|}} + \frac{{\Delta _{x'} }}{{\left| {x'} \right|}} + \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}

    \delta _{x'y'}  = \delta _{x'}  + \delta _{x'}  + \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}

    That's how I tried to prove but I don't know what to do with \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by OReilly
    Can someone show me the proof that relative error of product of approximant numbers is equal to sum of relative errors of each factors?

    I need to show that \delta _{x'y'}  = \delta _{x'}  + \delta _{y'}


    I know that
    \left| {xy - x'y'} \right| = \left| {x'} \right|\Delta _{y'}  + \left| {y'} \right|\Delta _{x'}  + \Delta _{x'} \Delta _{y'}

    If we divide both sides with \left| {x'y'} \right| we get that
    \frac{{\left| {xy - x'y'} \right|}}{{\left| {x'y'} \right|}} = \frac{{\left| {x'} \right|\Delta _{y'} }}{{\left| {x'y'} \right|}} + \frac{{\left| {y'} \right|\Delta _{x'} }}{{\left| {x'y'} \right|}} + \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}
    \frac{{\Delta _{x'y'} }}{{\left| {x'y'} \right|}} = \frac{{\Delta _{y'} }}{{\left| {y'} \right|}} + \frac{{\Delta _{x'} }}{{\left| {x'} \right|}} + \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}

    \delta _{x'y'}  = \delta _{x'}  + \delta _{x'}  + \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}

    That's how I tried to prove but I don't know what to do with \frac{{\Delta _{x'} \Delta _{y'} }}{{\left| {x'y'} \right|}}
    You throw it out as being of second degree in relative error, and so
    if the relative errors are relativly small this is negligable in comparison.

    RonL
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