Fractional Errors: When to Use Raw vs. Fractional Values?

[aspodkfpo]

Homework Statement

Why do fractional errors have a different error value from subbing in the raw values?

Relevant Equations

% + % = %

Why do fractional errors have a different error value from subbing in the raw values?
e.g. 10 +- 1 divided by 10 +- 1

fractional error yields 20%
11/9 - 9/11 yields 40/99

In cases like these do we use the original values and attempt to find the maximum error distance or use fractional errors?

 

[haruspex]

e.g. 10 +- 1 divided by 10 +- 1

fractional error yields 20%
11/9 - 9/11 yields 40/99

Are you comparing fairly?
Fractional error is ±20%, so a range of 40%. Not much different from 40/99.

 

[aspodkfpo]

Are you comparing fairly?
Fractional error is ±20%, so a range of 40%. Not much different from 40/99.

But the true difference is 40/99 right? It's not fair, but I'm just trying to resolve my qualms about which gets a more accurate uncertainty.

 

[haruspex]

But the true difference is 40/99 right?

Sure, but very few will care about a 1% error in what is already an error percentage.

 

[kuruman]

Suppose you have measured $P\pm \delta$. One usually writes the fractional uncertainty as the difference between the two extreme fractional discrepancies from the value $P$, $$\frac{\Delta P}{P}=\frac{P+\delta}{P}-\frac{P-\delta}{P}=\frac{2\delta}{P}.$$You write it as $$\frac{\Delta P}{P}=\frac{P+\delta/2}{P-\delta/2}-\frac{P-\delta/2}{P+\delta/2}=\frac{(P+\delta/2)^2-(P-\delta/2)^2}{P^2-\delta^2/4}=\frac{2P\delta}{P^2-\delta^2/4}=\frac{2\delta}{P[1-\delta^2/(4P^2)]}$$This last expression is exactly equivalent to your expression. We expect $\delta$ to be at least one order of magnitude smaller than $P$ in which case $\delta^2/(4P^2) <<1$ which means that we can ignore it in the denominator. Therefore, your way of writing the fractional difference matches to a good approximation the more common way.

Also, think about your statement

$\dots~$but I'm just trying to resolve my qualms about which gets a more accurate uncertainty.

What is an "accurate uncertainty"? If $P$ has a continuous spectrum of values, there is no clear boundary separating the values that could be assigned to it from the values that definitely cannot be assigned to it.

 

[aspodkfpo]

Suppose you have measured $P\pm \delta$. One usually writes the fractional uncertainty as the difference between the two extreme fractional discrepancies from the value $P$, $$\frac{\Delta P}{P}=\frac{P+\delta}{P}-\frac{P-\delta}{P}=\frac{2\delta}{P}.$$You write it as $$\frac{\Delta P}{P}=\frac{P+\delta/2}{P-\delta/2}-\frac{P-\delta/2}{P+\delta/2}=\frac{(P+\delta/2)^2-(P-\delta/2)^2}{P^2-\delta^2/4}=\frac{2P\delta}{P^2-\delta^2/4}=\frac{2\delta}{P[1-\delta^2/(4P^2)]}$$This last expression is exactly equivalent to your expression. We expect $\delta$ to be at least one order of magnitude smaller than $P$ in which case $\delta^2/(4P^2) <<1$ which means that we can ignore it in the denominator. Therefore, your way of writing the fractional difference matches to a good approximation the more common way.

Also, think about your statement
What is an "accurate uncertainty"? If $P$ has a continuous spectrum of values, there is no clear boundary separating the values that could be assigned to it from the values that definitely cannot be assigned to it.

Wait how is 10+/-1 with 11/9 = (P+d/2)/(P-d/2). Why not P+d/P-d

 

[kuruman]

Wait how is 10+/-1 with 11/9 = (P+d/2)/(P-d/2). Why not P+d/P-d

Because that overestimates the uncertainty by a factor of 2. Do the algebra and you will see. I really don't see why you do the fractional uncertainty the way you do. The logic for the usual way is intuitive and the algebra is simple:

Step 1. Subtract the lowest value from the highest value to get a measure of the difference $\Delta p$ between the two extreme values.
$\Delta P=(P+\delta)-(P-\delta)=2\delta.$

Sep 2. Express this difference as a fraction of the measured value to get the fractional difference.
$f=\dfrac{2\delta}{P}.$

Step 3 (optional). Multiply by 100 and tack on a percent sign to write it as a percentage.
$f_{\%}=\dfrac{2\delta}{P}\times 100\%.$

 

[aspodkfpo]

Because that overestimates the uncertainty by a factor of 2. Do the algebra and you will see. I really don't see why you do the fractional uncertainty the way you do. The logic for the usual way is intuitive and the algebra is simple:

Step 1. Subtract the lowest value from the highest value to get a measure of the difference $\Delta p$ between the two extreme values.
$\Delta P=(P+\delta)-(P-\delta)=2\delta.$

Sep 2. Express this difference as a fraction of the measured value to get the fractional difference.
$f=\dfrac{2\delta}{P}.$

Step 3 (optional). Multiply by 100 and tack on a percent sign to write it as a percentage.
$f_{\%}=\dfrac{2\delta}{P}\times 100\%.$

Still don't see why you were writing p+d/2 / p-d/2 when it is 11/9 for 10+- 1 (where I assume 1 is d). Largest possible value would be p+d/p-d, would it not?

 

[jbriggs444]

Still don't see why you were writing p+d/2 / p-d/2 when it is 11/9 for 10+- 1 (where I assume 1 is d). Largest possible value would be p+d/p-d, would it not?

If d is the "diameter" of the interval within which the true value will lie then d=2.

If one is trying to compute the diameter of the interval for the quotient, one should use as input the diameter of the intervals for the numerator and denominator. Try to avoid confusion. Use only one way of expressing error at a time.