Propagation of Error/Uncertainty

[luma]

I'm trying to get an intuitive sense for errors and picked some random numbers:

x = 2.5 +/- 0.01
find f(x) = x³

d f(x) / dx = 3x²
d f(x) = 3x² dx
= 3(2.5)² 0.01
= 0.1875

What I don't get is why f(x - Δx) ≠ f(x) - 0.1875 and why f(x + Δx) ≠ f(x) + 0.1875

Where did I go wrong in my method for finding the uncertainty value? Thanks

 

[luma]

I found a webpage which lists:

$$Q = a^n \Rightarrow \frac{\Delta Q}{Q} = |n| \frac{\Delta a}{a}$$

Applying the method I used in OP here,

$$\frac{\delta Q}{\delta a} = n a^{n-1}\\<br /> \delta Q = na^{n-1}\delta a\\<br /> \therefore \frac{\delta Q}{Q} = \frac{na^{n-1}\delta a}{a^n} = n \frac{\delta a}{a}$$

So it looks like my method of deriving the uncertainty is correct.

Working out f(x) = x^6; x = 25 +/- 1 I get do,

f(25 - 1) = f(24) = 191102976
f(25 + 1) = f(26) = 308915776

now using the identity df(x) = 6 25^5 * 1 = 58593750

25^6 + 58593750 ≠ (25+1)^6
25^6 - 58593750 ≠ (25-1)^6

Any ideas?

 

[HallsofIvy]

Yes, you can think of a small error as an "differential" and approximate errors in functions of the measurement using the derivatives.

If $y= x^2$ then $dy/dx= 2x$ so that $dy= 2xdx$. You could also say, then, that
$$\frac{dy}{y}= \frac{2xdx}{y}= \frac{2xdx}{x^2}= 2\frac{dx}{x}$$
so that the "relative error", the actual error in the measurement divided by the measurement, is multiplied by 2.

More generally, if f(x,y)= xy, where x and y are independent measurements, then $df= ydx+ xdy$ and so
$$\frac{df}{f}= \frac{ydx+ xdy}{xy}= \frac{dx}{x}+ \frac{dy}{y}$$.

This is equivalent to the old engineering "rule of thumb": "When measurements are added, their errors add, when measurements are multiplied, their relative errors add".

 

[luma]

Thank you for your informative post.

But what I still don't get is what they are showing?

If I make a measurement of x = 2 +/- 1 otherwise written as x = [1,3]

Then y = x² would be [1,9]

Using the rule derived dy = y (2 dx / x) = 2 x dx = 2*2*1 = 4

y = 2² +/- 4 = [0,8] ≠ [1,9]

So what does this represent intuitively?

Is this only an approximation? It doesn't look like it should be

 

[Office_Shredder]

Using the derivative is basically a way of saying "We're going to assume our function really looks like a line, and use the slope of that line to figure out what the error is".

The bigger your interval, the more room for error as the tangent line becomes a worse and worse approximation.

Notice that [0,8] is the range of the tangent line at x=2 over the interval[1,3]

 

[luma]

Thank you! Of course...