Error propagation when you take the inverse?

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When taking the inverse of a value with an associated error, the percentage error remains consistent. For example, with a voltage of 30 V ± 0.05 V, the inverse is calculated as 1/V, resulting in approximately 0.0333 V^{-1}. The error in this inverse value is also about 0.1667%, translating to an uncertainty of approximately 0.0000556 V^{-1}. The formula for uncertainty in functions of one variable indicates that the error does not change when taking the inverse. For further understanding, Taylor's "An Introduction to Error Analysis" is recommended.
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Say something is a value +/- .05. What happens when you take the inverse of the value? For example, 30 V +/- .05 V. 1/V...what would the error be?
 
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This is a math question. 1/(x+y)=1/(x(1+y/x)).=.(1/x)(1-y/x)=1/x-y/x2.

The assumption is|y|<<|x|, .=. means approx =

I'll let you do the arithmetic.
 
When you take the inverse, use % error. That is the same for the inverse as for the original.
 
Sorry, I have the same qns but i don't get what both of you are saying, elaborate with example? thanks
 
In the original question, the error in V is 0.05 V or (0.05/30)*100% = 0.1667%.

1/V = 0.0333 V^{-1}. The error in this is also 0.1667%, or about 0.0000556 V^{-1}.
 
The uncertainty in any function of one variable is \delta y = \left|\frac{dy}{dx}\right| \delta x. If y = x^n (in your case n = -1), then \frac{\delta y}{|y|} = |n| \frac{\delta x}{|x|}. For your case, the error is unchanged.

Taylor's book "An introduction to error analysis" is well worth reading.
 
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