# Propagation of uncertainties

1. Oct 15, 2008

### Oerg

1. The problem statement, all variables and given/known data
Given that $$f=\frac{\bar u \bar v}{\bar u +\bar v}$$

show that

$$e_f=f^2({\frac{e_u}{\bar u^2} + \frac{e_v}{\bar v^2})$$

where $$e$$ refers to the error. ok so I added up the fractional uncertainties

2. Relevant equations

3. The attempt at a solution
and I got this

$$\frac{e_f}{f}=\frac{e_u}{u}+\frac{e_v}{v}+\frac{e_u+e_v}{u+v}$$

after some simplifying, I got to this,

$$e_f=f^2(\frac{e_u(u+v)}{u^2v}+\frac{e_v(u+v)}{v^2u}+\frac{e_u+e_v}{uv})$$

and then I realized that I could never get the answer, however, if this term was negative,
$$\frac{e_u+e_v}{uv}$$, i would get the answer perfectly, but how can it be negative??? Problem is even in division, shouldn't the fractional uncertianties add up??

2. Oct 15, 2008

### Oerg

any help?

3. Oct 16, 2008

### Oerg

e_f is the maximum error, I have no idea why the term should be negative

4. Oct 16, 2008

### atyy

f(u,v)

df=(df/du)du+(df/dv)dv

I didn't notate it properly, but the derivatives are partial derivatives. Treat df as ef, du as eu, dv as ev

5. Oct 16, 2008

### Oerg

Im sorry, I dont understand your post. how is my working affected by this equation?

6. Oct 16, 2008

### atyy

I don't follow your working at all, so I'm not sure if it's a right way that I don't know about.

Are you familiar with differentiation?

7. Oct 16, 2008

### Oerg

yeh but not with partial derivatives,

What I did was to add up the fractional uncertainties.

I tried evaluating your equation, but I arrived at an ugly equation that doesnt fit the final bill.

8. Oct 16, 2008

### atyy

OK, that's good. Partial differentiation is exactly like ordinary differentiation. So for examle df(u,v)/du just means normal differentiation of f with respect to u, holding v constant. It may look ugly, but it should work out.

So first evaluate df(u,v)/du.

9. Oct 16, 2008

### Oerg

ok, I tried working out the coefficient of e_u according to your equation. Which is the partial derivative of f wrt u.

$$\frac{df}{du}=\frac{-uv}{(u+v)^2}+\frac{v}{u+v}$$

unfortunately, it evaluaes out to be

$$f \times (\frac{v}{u(u+v)})$$

which isn't what the question asked for. Have I differentiated wrongly?

10. Oct 16, 2008

### atyy

Try a different rearrangement, maybe take the common denominator to be (u+v)2

11. Oct 16, 2008

### Oerg

I have already combined the two terms in my previous post, but i got $$f \times (\frac{v}{u(u+v)})$$ for the coefficient of $$e_u$$ when it should have been
$$\frac{f}{u^2}$$

12. Oct 16, 2008

### Oerg

ok sorry, i got the answer, i have to factorise another f out again. THANKS FOR YOUR TIME, IM VERY VERY GRATEFUL!!!!

13. Oct 16, 2008

### atyy

Wait - you have to do it for df(u,v)/dv also!

14. Oct 16, 2008

### Oerg

yeh it's the same for the other term, it should evaluate out to be the same.

15. Oct 16, 2008

### atyy

And you understand what I meant in #4?

16. Oct 16, 2008

### Oerg

yeh i do,

df=partial derivative of f wrt u times du + partial derivative of f wrt v times dv

\

from df du and dv to ef eu and ev is an approximation for a small change.

17. Oct 16, 2008

### atyy

OK, great (assuming what I said is correct )