How Do You Calculate Error Uncertainty for ln(x)?

[johnnnnyyy]

Homework Statement

What would be the error uncertainty when you take ln of a number. For example ln(10) and the error uncertainty for 10 is ± 1

Homework Equations

The Attempt at a Solution

Is the error uncertainty just (1/10)*2.3? (2.3 is the answer to ln(10))

 

[rude man]

d/dx ln x = 1/x
what is d(ln x) in terms of dx?
That's for small changes in x.

For larger changes in x, say x → x+a, you can Taylor-series-expand ln (x + a). The first term is of course the above = ln x + a/x , then the higher terms give you better accuracy.

In your case, x = 10 and a = 1, the small-change approximation above gets you to within 0.9980 of the correct answer. Adding just 1 extra term in the series gets you to within 0.9999 of the exact answer. Etc.

 

[NascentOxygen]

Homework Statement

What would be the error uncertainty when you take ln of a number. For example ln(10) and the error uncertainty for 10 is ± 1

So the correct answer lies between ln(9) and ln(11)?

 

[rude man]

So the correct answer lies between ln(9) and ln(11)?

Those are the extremes, yes, but the fractional error distributions differ slightly between x and ln(x).

For small δx/x the fractional change in ln(x) is δ[ln(x)]/ln(x)] ~ [1/ln(x)] δx/x = 0.4343.

E.g. for δx = +1 and x = 10,
fractional change in ln(x) = [ln(11) - ln(10)]/ln(10) = 0.04139 for fractional change in x = 1/10 = 0.1000. So δln(x)/ ln(x) = 0.4139 δx/x.


but if δx = 0.1, fract. change in ln(x) is [ln(10.1) - ln(10)]/ln(10) = 0.004321 for a fractional change in x of 0.1/10 = 0.0100. So δln(x)/ln(x) = 0.4321 δx/x, pretty close to 0.4343.

Which as I said is not a big difference. So with little error, you can say that the fractional error in ln(x) is proportional to the fractional error in x, that ratio being 0.4343, if |δx/x| < 0.1.

But say δx = 5, then
fract. change in ln(x) = [ln(15) - ln(10)]/ln(10) = 0.1761 for δx/x = 0.5, so that ratio is 0.1761/0.5 = 0.3522 which is quite a ways from 0.4343.