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I think I proved the following but I am not sure. I will write my answer at the bottom

Is the set of logarithms $ \lbrace\ln (t + a_i)\rbrace_{i=1}^N $ with $t,a_i>0$, and all $a_i$ different linearly independent?

$N$ is a given integer.

Edit: My solution is the following:

The condition for linear dependence is that the function

\begin{equation} f(t) = \sum_{i=1}^N c_i \ln (t + a_i) \equiv 0 \end{equation} for some $c_i$ not all zero.

The function can be rewritten into \begin{equation} f(t) = \ln \left(\prod_{i=1}^N (t + a_i)^{c_i} \right) \equiv 0 \end{equation} which implies that the function \begin{equation} g(t) = \prod_{i=1}^N (t + a_i)^{c_i} \equiv 1 \end{equation} note that $g(t)$ can never be zero due to the hypothesis on $t$ and $a_i$.

Taking derivatives with respect to time we obtain \begin{equation} g'(t) = \left( \sum_{i=1}^{N}\frac{c_i}{t+a_i}\right)\prod_{i=1}^N (t+a_i) ^{c_i} = \sum_{i=1}^{N}\frac{{c}_{i}}{t+{a}_{i}} g(t) \equiv 0 \end{equation} Since $g(t)$ is never zero (and always one) we can remove it and obtain the equality \begin{equation} \sum_{i=0}^{N}\frac{{c}_{i}}{t+{a}_{i}} \equiv 0 \end{equation} Note that this states that translations of negative powers (of degree -1) are also linearly dependent.

Using the idea in this answer we can remove the denominators obtaining \begin{equation} \sum_{i=1}^N {c_i}\prod_{j\neq i}(t + a_j) \equiv 0. \end{equation}

The left-hand side of that equation defines a polynomial of degree $N-1$ in the variable $t$. Evaluating it at $N$ different values of $t$ produces a contradiction, since a polynomial of degree $N-1$ cannot have $N$ roots. Hence all $c_i$ must be zero.

Therefore

a) Translations of negative powers (of degree -1) are linearly independent.

b) $g(t) \not\equiv 1$.

c) The given logarithms are linearly independent.

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    $\begingroup$ It would be better if you gave your solution and we told you wether it is correct or not. $\endgroup$ – Pedro Tamaroff Aug 18 '12 at 0:19
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    $\begingroup$ The assertion is true, by applying the linear map of taking derivative, and using the uniqueness of partial fraction expansions of rational functions, which is immediate from the fact that $\mathbb R[t]$ is a PID. Ok., as @PeterTamaroff observes, the correctness of an unknown proof of a true thing is hard to gauge. (On other days, I have happily thought that it is lucky that false proofs of true things do not make them false.) $\endgroup$ – paul garrett Aug 18 '12 at 0:43
  • $\begingroup$ I can't post my answer yet (must wait 7 hours). But I basically followed the steps suggested by paul garrett. $\endgroup$ – JuanPi Aug 18 '12 at 0:45
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They are for $N=2$. I will show that if they are linearly dependent then $a_1 = a_2$ and $b_1 = -b_2$.

Exponentiating, and making all indexed variables distinct, this is $1 = (t+a)^b (t+c)^d$.

Looking at large $t$, this implies that $b$ and $d$ have opposite signs. Let $b$ be the positive one, and let $D = -d$. This becomes $(t+a)^b = (t+c)^D$ with $b$ and $D$ positive.

Taking the $b$-th root (or looking at large $t$), we find that $b = D$ and, then, $a = c$.

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  • $\begingroup$ I haven't seen an accepted answer with a downvote and no upvotes before. I would suggest that it is too soon to accept this answer, which is partial, but it certainly gives a useful way to think about the problem. +1 from me (solving this problem). $\endgroup$ – Ross Millikan Aug 18 '12 at 3:21

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