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I am trying to prove that the set $\left\lbrace\frac{1}{n+x}\right\rbrace_{n \in \mathbb{N}}$ is linear independant in the Vector space of functions from $\mathbb{R}_{>0}$ $\to$ $\mathbb{R}$. So starting with the linear combination $\lambda_1\frac{1}{1+x} +\cdots +\lambda_{k+1}\frac{1}{k+1+x}$ ,the idea was to convert it to a common denominator. Thus the polynomial in the counter is of degree k, having a maximum of k roots. This implies that the polynomial in the counter is the zero polynomial.

My problem so far is, that the only information i have about the system of equations is the first one $\lambda_1+\cdots +\lambda_{k+1}=0$. Does anyone know how to show that all lambdas are zero?

zkutch
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mane
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  • The step "$x = i$" in Lucas' answer below gives you a general recipe in proving linear independence of a set of functions. Basically, equality of functions means that at each $x$, equality holds as numbers. So, if one can substitute values of $x$ which are well-chosen, one forces a set of equations which imply $\lambda_i = 0$ for all $i$. Lucas' choice of $x$ ensured that very few $\lambda_i$ (in fact only one) was left behind. – Sarvesh Ravichandran Iyer Aug 09 '20 at 05:31
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    If you add the fractions by getting a common denominator, the numerator will be a polynomial of degree $\le k$ and thus has at most $k$ roots, in which case plugging in anything else will result in something nonzero. – anon Aug 09 '20 at 05:31

2 Answers2

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Notice that the sum needs to be the zero constant function. So: $$ \lambda_1 \frac{1}{1+x}+\cdots + \lambda_n \frac{1}{n+x} = 0 \Rightarrow \sum_{i=1}^n \lambda_i\prod_{j\neq i}(j+x) = 0 $$

Taking $x = -i$, we have $$ \lambda_i\prod_{j\neq i}(j-i) = 0 \Rightarrow \lambda_i = 0 $$ since $\prod_{j\neq i}(j-i) \neq 0$.

Lucas Resende
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    Note $x=-i$ is not in the original domain of the functions (nor can the original functions' domain be extended to include it). So there are some technical arguments under the hood. – anon Aug 09 '20 at 05:35
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    Well noticed! But since it is a polynomial function it is fully defined by a finite number of points and therefore we have an unique extension of the polynomial to the real line. That extension must be zero, so, the original function must be zero. – Lucas Resende Aug 09 '20 at 05:51
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The linear independence can be investigated through the Wronskian $$ W_n(x)=\begin{vmatrix} +\frac{1}{1+x} & +\frac{1}{2+x} & \ldots & +\frac{1}{n+x} \\ -\frac{1}{(1+x)^2} & -\frac{1}{(2+x)^2} & \ldots & -\frac{1}{(n+x)^2} \\ \ldots & \ldots & \ldots & \ldots \\ \frac{(-1)^{n+1}}{(1+x)^n} & \frac{(-1)^{n+1}}{(2+x)^n} & \ldots & \frac{(-1)^{n+1}}{(n+x)^n} \end{vmatrix} $$ The calculation of the Wronskian seems to be (by various tries with Wolfram Mathematica) $$ W_n(x)=\frac{a_n}{(1+x)^n(2+x)^n\cdot\ldots\cdot(n+x)^n} $$ where $a_n$ is the OIES Sequence A055209, so it is different from zero and the functions are then independent.

I leave to someone clever than me the proof that the Wronskian is effectively given by the previous expression.

Vincenzo Tibullo
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