Let $r: I \subseteq\mathbb{R}\to \mathbb{R}^n$ with I interval be a $C^k$ function such that $r'(t),...,r^{(k)}(t)$ are linearly independent for every $t\in I$, define $V_i(t) = r(t) + Span(r'(t),...,r^{(i)}(t))$ for $i\le k$.
Prove that the definition of $V_i$ does not depend on the parametrization of r, that is if $\gamma =r(\phi)$ with $\phi:J\subseteq \mathbb{R} \to I$ (J interval) bijective such that $\gamma$ is $C^k$ and $\phi'(s)\ne0,\forall s \in J$, then $V_i(\phi(s))= \gamma(s)+Span(\gamma'(s),...,\gamma^{(i)}(s))$.
I have some trouble proving this, for $k\gt3$ the explicit expression of $\gamma^{(k)}$ is not very useful; I've tried induction, but I'm stuck on the inductive step.
Could you give me a hint? Thanks
Edit:
I proved that $\gamma^{(i)}=\sum_{j=1}^{i}\alpha_j r^{(j)}$ for some $\alpha_j$ with $\alpha_i=(\phi')^i$.
Now the case $i=1$ is immediate. If we assume that $V_i = \gamma(s)+Span(\gamma'(s),...,\gamma^{(i)}(s))$ we have $Span(r'(t),...,r^{(i+1)}(t))= Span(\gamma'(s),...,\gamma^{(i)}(s), r^{(i+1)}(\phi(s))) = Span(\gamma'(s),...,\gamma^{(i+1)}(s))$ because $\gamma^{(i+1)}$ is a linear combination of $\gamma^{(j)}$, $1\le j\le i$, and $r^{(i+1)}$ and the coefficient of the last one is $\ne0$. We can conclude that $V_{i+1}(\phi(s))= \gamma(s)+Span(\gamma'(s),...,\gamma^{(i+1)}(s))$ for all $s \in J$.
Edit':
I proved that for every integer $k \ge 1$ we have $\gamma^{(k)}(s)=\sum_{j=1}^{k}\alpha_j(s)r^{(j)}(t)$ where $t=\phi(s)$, moreover $\alpha_k(s)=(\phi(s)')^k$ (of course $\alpha_j$ depends also on $k$).
If $k=1$ we have $V_1(t)=r(t)+Span(r'(t))$ and $\gamma(s)+Span(\gamma'(s))=r(t)+Span(\phi'(s)r'(t))$, since $\phi'(s)\ne 0$ we have the equality $V_1(t)= \gamma(s)+Span(\gamma'(s))$.
Let $k\ge 1$ be an integer and assume the thesis holds for it. We need to prove $V_i(t)= \gamma(s)+Span(\gamma'(s),...,\gamma^{(i)}(s))$ for $1\le i \le k+1$; if $1\le i\le k$ the equality follows by induction.
If $i=k+1$ then $\gamma^{(k+1)}(s)=\sum_{j=1}^{k}\alpha_j(s)r^{(j)}(t)+(\phi'(s))^{k+1}r^{(k+1)}(t) \in Span(r'(t),...,r^{(k+1)}(t))$, but $\gamma^{(k+1)}(s) \notin Span(\gamma'(s),...,\gamma^{(k)}(s))$ and $Span(r'(t),...,r^{(k)}(t))=Span(\gamma'(s),...,\gamma^{(k)}(s))$.
Therefore $Span(\gamma'(s),...,\gamma^{(k+1)}(s)) \le Span(r'(t),...,r^{(k+1)}(t))$ and they have the same dimension, hence equality holds (or we can use symmetry).