Finding error using differentials

[ProPatto16]

Homework Statement

Four positive numbers, each less than 50, are rounded to the first decimal place and then multiplied together. use differentials to estimate the maximum possible error in the computed product that might result from the rounding.

The Attempt at a Solution

i need an equation to differentiate. but from the question all i can gather is that the error range is 0.05 units. and the equation has to be something like wxyz = ? but that's just silly.

where do i start?

i know i need to prtially differentiate the equation by all variables then multiply each deifferential by error 0.05. but i need an equation to begin with and i got no idea.

 

[ProPatto16]

nevermind got it.

 

[Polyamorph]

f=w*x*y*z. So σf = [((df/dw)σw)^2+((df/dx)σx)^2+((df/dy)σy)^2+((df/dw)σz)^2]^1/2.

 

[HallsofIvy]

You don't need an equation, you need a function. f= wxyz so
df= xyz dw+ wyz dx+ wxz dy+ wxy dz

Since each is rounded to one decimal place, so that "3.1" might mean anything from 2.5 to 3.5, a difference of 0.1, dw= dx= dy= dz= 0.1. And, we are told, the largest that w, x, y, and z could be is 50.

 

[Polyamorph]

"df= xyz dw+ wyz dx+ wxz dy+ wxy dz" is incomplete.

It should be df= [(xyz dw)^2+ (wyz dx)^2+ (wxz dy)^2+ (wxy dz)^2]^1/2 per my post above.

 

[Mark44]

f=w*x*y*z. So σf = [((df/dw)σw)^2+((df/dx)σx)^2+((df/dy)σy)^2+((df/dw)σz)^2]^1/2.

No, df = xyz*dw + wyz*dx + wxz*dy + wxy*dz, just as HallsOfIvy has it below.

Also, why are you using sigma (σ) in the differentials?

You don't need an equation, you need a function. f= wxyz so
df= xyz dw+ wyz dx+ wxz dy+ wxy dz

Since each is rounded to one decimal place, so that "3.1" might mean anything from 2.5 to 3.5, a difference of 0.1, dw= dx= dy= dz= 0.1. And, we are told, the largest that w, x, y, and z could be is 50.

Minor correction: 3.1 would be any number between 3.05 and 3.15.

 

[Polyamorph]

σ is the error. since d is the symbol used for differentials it is always better not to confuse the two.

The generic formula to compute the error in the function f(wxyz) is:

σf = [((df/dw)σw)^2+((df/dx)σx)^2+((df/dy)σy)^2+((df/dw)σz)^2]^1/2

See http://books.google.fr/books?id=giF...g=PA210#v=onepage&q=error propagation&f=false

equation 9.2.

It is exactly the same equation as I have used!

The formalism used by HallsOfIvy is an approximation, see equation 9.1 of the same reference.

 

[Mark44]

Polyamorph,
The formula you show gives the standard deviation of the error, which is not what was asked for. In ProPatto16's OP, it says "use differentials to estimate the maximum possible error" (emphasis added).

 

[Polyamorph]

Error analysis is always an estimate. I think the point to emphasise though is that the OP needed the maximum possible error.

So yes you're right, I concede that HallsofIvy gives the correct formula for the maximum error since the formalism I provided will always be less than or equal to the result given by HallsofIvy.

 

[ProPatto16]

further investigation into the question has yielded the following.

the maximum possible error, that is the maximum number of units the product may be incorrect by due to rounding is 25,000 units.

since 50^4 = 6,250,000 then the error is only 0.4% which is reasonable.

the equation given by poly yields an answer of sq rt of 25000 if I've done it correctly. which gives approx 158.11. when compared to 6.25 mil its completely negliable.

apologies for the disagreements.

 

[Polyamorph]

That result is strange because you would be saying ((df/dw)σw)²+((df/dx)σx)²+((df/dy)σy)²+((df/dw)σz)²= xyz σw+ wyz σx+ wxz σy+ wxy σz = 25000 which is clearly false, perhaps you forgot to square the partials?

The maximum error in each variable is actually 0.1. So using HallsofIvy formula gives an error of: 4*(50*50*50*0.1)=50,000 (0.8% of 6,250,000)

If you needed a more accurate estimate of the error (as opposed to just the maximum error) then you can use the formula I showed you:
[4*((50*50*50*0.1)²)]^1/2 = 25,000

So it is half the the maximum value and gives a percentage error of 0.4%. No good for your homework but useful if you are taking accurate measurements and need a more reliable error estimate.

Feel free to correct me if I'm wrong but I think your final answer should be 50,000, not 25,000.