What is the formula for Δ(x²/y) measurement error
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What is the formula for Δ(x²/y) measurement error

[From: ] [author: ] [Date: 12-05-22] [Hit: ]
..Fractional error in x is 0.0005/0.249= 0.Fractional error in y is 0.......
If I have x with error ∆x and y with error ∆y, how to find Δ(x²/y)?

If I had z=x²/y that would be Δz/z = Δx/x + Δy/y, but without =, I have no idea....

-
Δz/z = 2* Δx/x + Δy/y
In other words the fractional error of the result is 2* Δx/x + Δy/y
Which means that the absolute error is then (2* Δx/x + Δy/y) * x^2 / y
OK now to answer
Fractional error in x is 0.0005/0.249= 0.002
Fractional error in y is 0.5/19 = 0.026

So the fractional error in the result is 0.002 * 2 + 0.026 = 0.030

And the absolute error is then 0.00325 +- (0.030 * 0.00325)
= 0.00325 +- 9.9 * 10 ^ -5
***************************************…
now I can give the short explanation or the long one.
In short, we add the percentage errors ( or fractional errors) whenever multiplying or dividing.
And as x^2 = x * x we must add the fractional error twice.

This "rule of thumb" is applicable when the error is a small fraction.


If you want to do the calculation the formal way, you must find both the maximum possible values and the minimum possible values.

ie max is (x + Δx)^2/(y - Δy)

and the minimum is (x - Δx)^2/(y + Δy)

If you expand each of these you will see how the rule comes about.

take the first (x + Δx)^2 / (y - Δy) = (x + Δx)*(x + Δx) * (y + Δy) / [(y - Δy) * (y + Δy)]
= (x + Δx)*(x + Δx) * (y + Δy) / [y^2 - Δy^2]
But as Δy is small compared to y then Δy^2 is insignificant compared to y^2

hence the formula becomes
= (x + Δx)*(x + Δx) * (y + Δy) / y^2
and by expanding the top line then dividing by y^2 you will see how the whole thing works out formally.
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