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I'm trying to prove that the functions:

$cot(x), cot(2x), \dots, cot(nx)$ are linearly independent.

My idea was to use mathematical induction, that is:

For $n = 1, \hspace{0.5cm} \alpha_1 cot(x) \equiv 0 \Leftrightarrow \alpha_1 = 0 $

Now suppose this holds for n,

$\alpha_1cot(x) + \alpha_2cot(2x) + \dots + \alpha_ncot(nx) \equiv 0 \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} \alpha_1 = \alpha_2 = \dots = \alpha_n = 0$

We have to prove that

$\alpha_1cot(x) + \alpha_2cot(2x) + \dots + \alpha_ncot(nx) + \alpha_{n+1} cot((n+1)x) \equiv 0 \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} \alpha_1 = \alpha_2 = \dots = \alpha_n = \alpha_{n+1} = 0 \hspace{0.4cm} \dots \dots (1)$

I tried to differentiate both sides and we get:

$$\sum_{k=1}^n -\alpha_k(1 + (cot(kx))^2) \equiv 0 \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} \sum_{k=1}^n \alpha_k(1 + (cot(kx))^2) \equiv 0 $$

However, now I'm not sure how to proceed. I would appreciate any suggestions for the way of my approach or any other easier way to prove it!

John
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  • $\alpha_1 cot(x) = 0$ does not imply that $\alpha_1 = 0$ $\cot{x}$ can also be zero – Shlok Jain Jun 20 '23 at 17:58
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    Hint: $\cot(kx)$ has a singularity at $\pi/k$. – Robert Israel Jun 20 '23 at 17:58
  • @User $\alpha_1 cot(x)$ is identically equal to 0, meaning this should hold for all x. – John Jun 20 '23 at 18:07
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    @RobertIsrael ouu, this technically solves it. I didn't think about that. All I have to do now is to move one of the terms $ \alpha_m cot(mx) $ to one side and the sum to the other, then since they have to be identically equal, one side has singularity at $ \frac{\pi}{m}$ and the other doesn't, this implies that all $\alpha_k$ should be 0. – John Jun 20 '23 at 18:13
  • John: @User 's confusion comes from your notation $\alpha_1 \cot(x) = 0$, which looks like it's saying that for some particular value $x$, we have the equality $\alpha_1 \cot(x) = 0$. A more usual notation for "identically equal to 0" would be $\alpha_1 \cot = 0$, or alternatively $\forall x,,\alpha_1 \cot(x) = 0$. – Stef Jun 20 '23 at 18:14
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    @John Not quite, because $ \cot(3mx) $ also has a singularity at $ (3 \pi ) / (3m)$, so we might have it on the other side. Do you see how to fix that gap? – Calvin Lin Jun 20 '23 at 18:19
  • It seems the Wronskian is not identically zero. But (computing up to $n=5$) I did not see a useful pattern to prove it. – GEdgar Jun 20 '23 at 18:20
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    @Stef I edited it, I hope that now it's clear. – John Jun 20 '23 at 18:21
  • @John Ah yes, $\equiv$ is good notation too – Stef Jun 20 '23 at 18:22
  • @CalvinLin perhaps we can let $\alpha_{3m} = 0 $ ? – John Jun 20 '23 at 18:24
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    We don't have control on $\alpha_{3m} $ right? Besides, there's still $ \cot (5m) , \cot (7m) \ldots $ which can contribute a singularity. – Calvin Lin Jun 20 '23 at 18:27
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    @CalvinLin Yes you're right. We can move to the other side the last term. So $\alpha_{n+1} cot((n+1)x)$ has singularity at $\frac{\pi}{n+1}$ and the sum on the other side doesn't. – John Jun 20 '23 at 19:03
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    Yup, if we take the largest $n$ term, then it gives a singularity that cannot be explained by the smaller terms. – Calvin Lin Jun 20 '23 at 19:24

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