# If a sum is 0, is the summand 0?

1. Nov 30, 2014

Hey guys,

Was just wondering something. Suppose I have an equation of the form

$\sum_{i=0}^{n}\frac{1}{x_{i}}(a-by_{i})=0$,

how would I solve this? do I just set the summand = 0?

2. Nov 30, 2014

### Staff: Mentor

No. If the terms in the sum can be positive or negative, their sum can be zero without any of them being zero. As a simple example, 2 + (-1) + (-1) = 0, but no single term equals zero.

Last edited: Nov 30, 2014
3. Nov 30, 2014

### Stephen Tashi

What variables are you solving for? What symbols represent known values ?

4. Nov 30, 2014

The real equation I'm dealing with is
$\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}2x(y_{i}-\alpha x -\beta x^{2})=0$

and im trying to solve for alpha and beta but one at a time...

5. Nov 30, 2014

### Stephen Tashi

So it's "$x$" instead of "$x_i$"?`

6. Nov 30, 2014

Yes I wrote it completely differently in the original post as I didnt think I would need to put the actual equation here but I changed my mind, sorry.

There is no summation over $x$, only over the $\sigma_{i}$ and $y_{i}$.

7. Nov 30, 2014

The closest I can get is

$(\alpha x +\beta x^{2})\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}=\sum_{i=1}^{n}\frac{y_{i}}{\sigma_{i}^{2}}$

No idea what to do from here

8. Dec 1, 2014

### Stephen Tashi

That equation doesn't have a unique solution. To get a unique solution, you need to know another equation involving $\alpha$ and $\beta$.

Does this equation come from setting a partial derivative equal to zero? If so, perhaps there is another partial derivative that's supposed to be set equal to zero.

9. Dec 1, 2014

Yes you are right - it turns out that the $x$ are also summed in addition to $y_{i},\sigma_{i}$, i had interpreted the situation inocorrectly.