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Urgent Numerical Methods Help Needed! Exam Tomorrow!! watch

    • Thread Starter

    I'm going through some last minute revision and while I feel prepared I've come across a past paper question that has got me stumped and panicking. If anyone could help I would be sooo grateful!

    The numbers X and Y shown below are known to be correct to 3 decimal places.
    X = 2.718 Y = 3.142
    (i) State the maximum possible errors in X, X + Y, X – Y, 10X + 20Y. [4]
    (ii) Find the maximum possible relative errors in X and Y. Hence state approximately the maximum
    possible relative errors in XY and XY

    • Thread Starter

    Doesn't matter now lol - being completely and utterly stupid for not getting this! lol

    max error in X = plus/minus 0.0005 as it is known that it is accurate to 3 decimal places, so there is uncertaintity about the fourth.

    X + Y: errors add in quadradure, so use formula error = sqrt((error on x)^2 + (error on y)^2)

    X - Y: errors still add despite it being a subtraction so same as X + Y

    10X + 20Y: Same as other two as 10 and 20 are simply multiples and hence have no associated error.

    to get relative errors, divide the error by the measured value

    e.g. rel error(X) = error(X) / X
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