# Error propagation when you take the inverse?

1. Feb 7, 2008

### homestar

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?

2. Feb 7, 2008

### mathman

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.

3. Feb 8, 2008

### pam

When you take the inverse, use % error. That is the same for the inverse as for the original.

4. Sep 8, 2011

### |\|a|\|

Sorry, I have the same qns but i don't get what both of you are saying, elaborate with example? thanks

5. Sep 8, 2011

### Staff: Mentor

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}.

6. Sep 8, 2011

### Andy Resnick

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook