# Error propagation - inverse of an error

1. Nov 17, 2016

### emeraldskye177

1. The problem statement, all variables and given/known data

Let t = f(g, h, A, Δm, Γ, r). For t = 2 s, the propagated error is σ = 0.02 s.

Can the error of 1/t2 be simply determined using the known error in t = (2 ± 0.02) s, or must the variance formula (with all the partial derivatives and errors of each dependent variable) be re-derived?

2. Relevant equations

When squaring a quantity with an error, the new %error of the square is half the old %error.

I don't know how to find the inverse of an error (tried looking it up online but can't seem to find it).

3. The attempt at a solution

t2 = (22 ± 1% / 2) s = (4 ± 0.5%) s

1/t2 = ?

Last edited: Nov 17, 2016
2. Nov 17, 2016

### haruspex

It's not hard to figure out. If the fractional error in x as a measure of X is ε then X is x(1+/-ε). So 1/X is 1/(x(1+/-ε))=(1/x)(1+/-ε)-1.
You can use the binomial theorem to expand (1+/-ε)-1 and find the fractional error in 1/X.

3. Nov 17, 2016

### emeraldskye177

So, by what you're saying,

t = 2 ± 0.02 = 2 ± 1% = 2 × ( 1 ± 0.5% ) ...or should this be 2 × ( 1 ± 1% ) i.e. do I factor the 2 out of the error?

t
2 = 4 ± 0.5% = 4 × ( 1 ± ( 0.5% / 4 ) ) ...or should this be 4 × ( 1 ± 0.5% ) i.e. do I factor the 4 out of the error?

1/t2 = 1 / ( 4 × ( 1 ± ( 0.5% / 4 ) ) ) = ( 4 ± 0.5% )-1

Last edited: Nov 17, 2016
4. Nov 17, 2016

### haruspex

Yes. 2±0.02=2(1±0.01)

5. Nov 17, 2016

### emeraldskye177

So

t = 2 ± 1% = 2 * (1 ± 1%)

t2 = 4 ± 0.5% = 4 * (1 ± 0.5%)

1/t2 = 0.25 * (1 ± 0.5%)-1

How do I binomially expand something to the power of -1?

6. Nov 17, 2016

### haruspex

7. Nov 17, 2016

### emeraldskye177

Sorry I just realized I was making a silly mistake. Since I was squaring t, I should've been doubling the %error (for some reason I was thinking of the rule for taking the square root, in which case the %error is halved). So you are correct; if t = 2 ± 1% then

t2 = 4 ± 2% = 4 * (1 ± 2%) = 4 * (1 ± 0.02)

I checked out the article... for (t2)-1 looks like I'm adding/subtracting powers of t for infinity? I don't know what to make of that. I'm studying computer engineering, I'm not that great at theoretical math.

Last edited: Nov 17, 2016
8. Nov 17, 2016

### haruspex

Yes, but the terms get rapidly smaller for small values of the error inside the (). You only need to use the first two, the '1' and the next term.
So (1+x)-1 is approximately 1-x.

9. Nov 17, 2016

### emeraldskye177

So, for |x| << 1,

(1 - x)-1 = 1 + x, correct?

So (1 ± x)-1 = 1 ± x

So in my example,

t2 = 4 ± 2% = 4 * (1 ± 2%)

1/t2 = 0.25 * (1 ± 2%)-1 = 0.25 * (1 ± 2%) = 0.25 ± 2% = 0.25 ± 0.005

Is my final answer (bolded) correct?

10. Nov 18, 2016

Yes.