If a sum is 0, is the summand 0?

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In the discussion, a user inquires whether a summation equaling zero implies that the summand must also be zero. It is clarified that a sum can equal zero even if none of the individual terms are zero, using an example to illustrate this point. The user then presents a more complex equation involving variables alpha and beta, seeking guidance on how to solve it. It is noted that the equation does not yield a unique solution without an additional equation involving alpha and beta. Ultimately, the user resolves their confusion after realizing the correct interpretation of the variables involved.
Dixanadu
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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?
 
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Dixanadu said:
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?
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.
 
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Dixanadu said:
how would I solve this?
What variables are you solving for? What symbols represent known values ?
 
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 I am trying to solve for alpha and beta but one at a time...
 
Dixanadu said:
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

So it's "x" instead of "x_i"?`
 
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}.
 
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
 
Dixanadu said:
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

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.
 
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.

Everything is fine now thank you for your help! :D
 

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