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I read this in a paper:

$$ \lambda_m = \mbox{const} \quad \mbox{for all} \quad m \in \left\{1,2,\cdots,M\right\} $$

Does this mean that all $\lambda_m$ are the same, or that they're all constant functions, but with different constants?

The context: http://i.stack.imgur.com/inyO7.png

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    Without context, this is impossible to answer.2012-05-30
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    It's from [Some Improvements in the Measurement of Variable Latency Acoustically Evoked Potentials in Human EEG](http://www.ncbi.nlm.nih.gov/pubmed/6862512). I'll supply some more context.2012-05-30
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    I'm not allowed to post images, so could someone edit this into the question? http://i.stack.imgur.com/ZIoIe.png2012-05-30
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    Usually this notation means that each $\lambda_m$ is constant, but the constants can be different. Otherwise the constant should be given a name, e.g. $\lambda_m = C$ where $C$ is constant, or they could just write $\lambda_1 = \cdots = \lambda_m = \text{const}$.2012-05-30
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    @Johan The image was really big and seemed to take over the question area in a confusing way, so instead of inlining it I just added a link.2012-05-30
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    @Mark: even so it is worth saving a click2012-05-30

1 Answers 1

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I agree with Yuval Filmus's comment both in the abstract and in the specific. That is, generally we would write something like $$ \lambda_n = \lambda_m \qquad \forall n,m\in\mathbb{N} $$ or $$ \lambda_1 = \lambda_2 = \cdots = \lambda_n = \cdots $$ or $$ \lambda_n = C \qquad \forall n \in\mathbb{N}$$ if we want it to mean that all of the $\lambda_n$ are equal.

This is also the case, I believe, in the context. Note that the author refers to functions $e_m(k) = e(k - \lambda_m)$ and states that

i.e., all $e_m(k)$ have identical shape and can differ only by a time-shift (latency) $\lambda_m$.

If all $\lambda_m$ were to be the same constant, there's hardly any point in defining $e_m$'s as different functions! Hence it is more natural to interpret the statement as requiring that $\lambda_m$ being independent of $k$, with the possibility that $\lambda_n\neq \lambda_m$.

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    Congratulations on hitting 20k. Finally you can be trusted.2012-06-01