What are the generators of invariant ring $K\left[\bigoplus_{k=1}^n V^{\bigodot k} \right]^G$, where $G$ is subgroup of $GL(V)$ with natural representation on $\bigoplus_{k=1}^n V^{\bigodot k} $i.e. for $g\in G$ is $g\cdot (v_1 \oplus (v_2 \odot v_3) \oplus \dots) = (gv_1 \oplus (gv_2 \odot gv_3) \oplus \dots) $.
$V^{\bigodot k}$ is $k$-th symmetric power of $V$.
Options:
For $K$ you can pick $\mathbb{R}$ or $\mathbb{C}$
As $G$ you can take $GL(V),O(V),SO(V)$.
I would like to have answer for all $n\in \mathbb{N}$.
Motivation: I came across this question.
Given two function $f,g : K^n \rightarrow K$. What is the condition on $f,g$ that $$\exists A\in GL(K^n) : f(Ax)-g(x) = \mathcal{O}(\|x\|^n)\qquad \text{as } x\rightarrow 0 $$
I can calculate invariants of tensors from Taylor expansion of $f$ and $g$ and compare them.
progress: I think I have result for $n=2$, $K=\mathbb{R}$, $G=GL(V)$ but with a little bit wonky proof.
Write elements of $\bigoplus_{k=1}^2 V^{\bigodot k} $ as $(A,v)$ where $A$ is symmetric matrix and $v$ is any vector. If $S\in G$ acts as $S\cdot (A,v)= (S^TAS,Sv)$. We will write $(A,v)\sim (B,u)$ iff there is such $S\in G$ that $S\cdot(A,v) = (B,u)$
Now we will modify $(A,v)$ to some canonical form and we will call this canonical from which we will deduce its invariants.
Since $A$ is symmetric, it can be written as $U^TAU = D$, $U\in O(V)$ and $D$ is diagonal(for now assume that it is non-degenerated ie does not have zero on diagonal) . $$ (A,v) \sim (U^TAU,Uv) = (D,Uv) \sim (|D|^{-\frac1{2}} D |D|^{-\frac1{2}}, |D|^{-\frac1{2}} Uv) = (D|D|^{-1},|D|^{-\frac1{2}} Uv) $$ $D|D|^{-\frac1{2}}$ is diagonal matrix with $k$ ones and $l$ minus ones. $D|D|^{-\frac1{2}}$ is invariant under action of $O(k,l)$. So we are left with vector $|D|^{-\frac1{2}} Uv$ on which we can act freely with elements of $O(k,l)$. The only invariant of vector under action of $O(k,l)$ is its length with respect to signature $(k,l)$. That is $$ (|D|^{-\frac1{2}} Uv)^T D|D|^{-1} |D|^{-\frac1{2}} Uv = v^TU^TD^{-1}Uv = v^TA^{-1}v $$
So the invariants of $(A,v)$ are signature of $A$ and $v^TA^{-1}v$ ie if $(A,v)$ and $(B,u)$ has same invariants then $(A,v)\sim (B,u)$.