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Error propagation - inverse of an error

1. Homework Statement

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. Homework 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 = ?
 
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haruspex

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1. Homework Statement

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. Homework 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 = ?
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.
 
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.
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

Did I follow you correctly?
 
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Yes. 2±0.02=2(1±0.01)
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?
 

haruspex

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No. How do you get that?
I think your ± X% notation is confusing you. Try writing it in factor form. Instead of mean value ± Percentage% write (mean value)(1± Percentage%). Thus, t=2(1± 0.01), so t2=4(1± 0.01)2=4(1± 0.02) approximately.
For negative powers, see e.g. https://en.m.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem
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.
 
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haruspex

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looks like I'm adding/subtracting powers of t for infinity?
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.
 
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.
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?
 

haruspex

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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?
Yes.
 

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