Uncertainty Formulas for High-School Level

[hobomoe]

(High-school level)
I got given some formulas today for calculating uncertainty of a value that has been square rooted, squared, inversed, inverse squared and inverse square rooted.
I'm not totally sure what they mean however, but I kinda have an idea.

Square root: y=x^1/2 Uncertainty=change in x/(2*x^1/2)
Square: y=x2 Uncertainty=2*change in x
Inverse: y=1/x Uncertainty=change in x/x^2
Inverse square: y=1/x^2 Uncertainty=(2*change in x)/x^3
Inverse square root: y=1/x^1/2 Uncertainty=change in x/(2*x*x^1/2)

Not sure how to show change in x with a symbol.

I take it that change in x means the uncertainty of value x and x is the raw value before it's squared or anything.

The second uncertainty read 2*x*change in x, but I was told that was wrong and the x didn't belong. I'm not sure if the others given to me are correct now.

Also, are these the only ways to calculate the uncertainties or are there easier methods?

 

[HallsofIvy]

(High-school level)
I got given some formulas today for calculating uncertainty of a value that has been square rooted, squared, inversed, inverse squared and inverse square rooted.
I'm not totally sure what they mean however, but I kinda have an idea.

Square root: y=x^1/2 Uncertainty=change in x/(2*x^1/2)
Square: y=x2 Uncertainty=2*change in x
Inverse: y=1/x Uncertainty=change in x/x^2
Inverse square: y=1/x^2 Uncertainty=(2*change in x)/x^3
Inverse square root: y=1/x^1/2 Uncertainty=change in x/(2*x*x^1/2)

Not sure how to show change in x with a symbol.

Most common is either "$\Delta x$" or "dx". By the way, mathematically, "inverse" is usually taken to mean "inverse function" (the inverse of squaring is the square root). What you mean by "inverse" is the reciprocal.

I take it that change in x means the uncertainty of value x and x is the raw value before it's squared or anything.

Yes, that is correct.

The second uncertainty read 2*x*change in x, but I was told that was wrong and the x didn't belong. I'm not sure if the others given to me are correct now.

No, if $y= x^2$ then $dy= 2x dx$. That, and the others, are correct.

Also, are these the only ways to calculate the uncertainties or are there easier methods?

What you are really doing is taking the derivative and calculating how the function can change when x changes: if y= f(x) then dy= f'(x)dx where f'(x) is the derivative of f with respect to x- it is defined as
$$\lim_{h\to 0} \frac{f(x+h)- f(x)}{h}$$

Notice the limit- strictly speaking these formulas are only approximations that are more and more accurate as the size of the change in x decreases. For example, if $y= x^2$
x changes by dx, then y becomes $(x+ dx)^2= x^2+ 2xdx+ (dx)^2$ so y has changed by $2xdx+ (dx)^2$. But if dx is very small, $(dx)^2$ will be much smaller and can be neglected: the change (error) is 2x dx.

It is always true that if $y= x^n$ then $dy= n x^{n-1}dx$. Your "rules" are all based on that with n= 1/2, 2, -1, -2, -1/2, respectively.

Two basic rules for the derivative, by the way, are the "sum rule", d(u+ v)= du+ dv, and the "product rule", d(uv)= u dv+ v du. These give rise to two Engineer's "rules of thumb": if quantities are added then their errors add, if quantities are multiplied then their relative errors add. The first is, of course, directly from "d(u+ v)= du+ dv". The second is from the product rule: d(uv)= u dv+ v du so if we divide both sides by uv we have
$$\frac{d(uv)}{uv}= \frac{u dv}{uv}+ \frac{vdu}{uv}= \frac{dv}{v}+ \frac{du}{u}$$.
The "relative error" is the error, du, dv, or d(uv), divided by the quantity u, v, or uv to give du/u, dv/v, or d(uv)/(uv).

 

[hobomoe]

I've read in places that you just times the uncertainty by two when you square x and divide it by 2 when you square root it :S