Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
6  
Without context, this is impossible to answer. –  Phira May 30 '12 at 15:03
    
It's from Some Improvements in the Measurement of Variable Latency Acoustically Evoked Potentials in Human EEG. I'll supply some more context. –  Johan May 30 '12 at 15:10
    
I'm not allowed to post images, so could someone edit this into the question? i.stack.imgur.com/ZIoIe.png –  Johan May 30 '12 at 15:14
5  
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 $\lambda_1 = \cdots = \lambda_m = \text{const}$. –  Yuval Filmus May 30 '12 at 15:22
1  
@Mark: even so it is worth saving a click –  Henry May 30 '12 at 17:35

1 Answer 1

up vote 5 down vote accepted

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 $\lambda_n\neq \lambda_m$.

share|improve this answer
    
Congratulations on hitting 20k. Finally you can be trusted. –  Asaf Karagila Jun 1 '12 at 7:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.