1 | initial version |

Let $f_1,\ldots,f_n$ be functions defined on a given interval $I$ and $n-1$ times differentiable. Let $W(f_1,\ldots,f_n)$ be the Wronskian of these functions. It is well known that:

if $f_1,\ldots,f_n$ are linearly dependent on $I$, then $W(f_1,\ldots,f_n)$ vanishes on $I$.

In general, the converse is not true: the vanishing of the Wronskian on $I$ does not implies that the functions are linearly dependent. However, if $f_1,\ldots,f_n$ are analytic, then these functions are linearly dependent on $I$ if and only if $W(f_1,\ldots,f_n)\equiv 0$ on $I$.

Since $f(x)=17$, $g(x)=\cos^2(x)$ and $h(x)=\cos(2x)$ are analytic on $\mathbb{R}$, we can check if they are linearly dependent by testing their Wronskian:

```
sage: f(x) = 17
sage: g(x) = cos(x)^2
sage: h(x) = cos(2*x)
sage: funs = [f(x),g(x),h(x)]
sage: bool(wronskian(*funs,x)==0)
True
```

Thus, $f$, $g$ and $h$ are linearly dependent.

2 | No.2 Revision |

Let $f_1,\ldots,f_n$ be functions defined on a given interval $I$ and $n-1$ times differentiable. Let $W(f_1,\ldots,f_n)$ be the Wronskian of these functions. It is well known that:

if $f_1,\ldots,f_n$ are linearly dependent on $I$, then $W(f_1,\ldots,f_n)$ vanishes on $I$.

In general, the converse is not true: the vanishing of the Wronskian on $I$ does not ~~implies ~~imply that the functions are linearly dependent. However, if $f_1,\ldots,f_n$ are analytic, then these functions are linearly dependent on $I$ if and only if $W(f_1,\ldots,f_n)\equiv 0$ on $I$.

Since $f(x)=17$, $g(x)=\cos^2(x)$ and $h(x)=\cos(2x)$ are analytic on $\mathbb{R}$, we can check if they are linearly dependent by testing their Wronskian:

```
sage: f(x) = 17
sage: g(x) = cos(x)^2
sage: h(x) = cos(2*x)
sage: funs = [f(x),g(x),h(x)]
sage: bool(wronskian(*funs,x)==0)
True
```

Thus, $f$, $g$ and $h$ are linearly dependent.

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