Is the identity ($f(x)=x$) a linear dependence? I would say yes but as usual I can't be absolutely sure and I could have misunderstood. I understand linear dependence but can it also be as simple as $y=x$, $y=2x$ and likewise? Is similarly $y=x^3$ a cubic dependence or have I misunderstood? When I look in wikipedia and research the concept it's more about vectors than about functions. 
 A: I am adding a good point here, although, I personally haven't heard or seen such that words (quadratic, cubic,...) in this area.

A set of functions  $f_1(x),f_2(x),\ldots,f_n(x), x\in I$ is linearly dependent on $I$ iff the determinant below is identically zero on $I$:
  $$ \det\left( \begin{array}{ccccc} \int_{a}^{b} f_1^2 \,dx& \int_{a}^{b} f_1f_2 \,dx&… &\int_{a}^{b}f_1f_n\,dx \\ \int_{a}^{b}f_2f_1\,dx & \int_{a}^{b}f_2^2\, dx &\ldots &\int_{a}^{b}f_2f_n\,dx \\ ⋮ & ⋮ & ⋮ &⋮ \\ \int_{a}^{b}f_nf_1\,dx & \int_{a}^{b}f_nf_2\,dx&\ldots &\int_{a}^{b}f_n^2\,dx \end{array} \right) $$

A: A linear dependence among an $m$-tuple $(f_1(x),\ldots,f_m(x))$ of functions is an $m$-tuple $(c_1,\ldots,c_m)\ne(0,\ldots,0)$ of scalars such that $c_1 f_1(x)+\cdots+c_m f_m(x)$ is the zero function (i.e. equal to $0$ regardless of what $x$ is).  Suppose we regard linear dependence as being expressed by the equality $\forall x\quad  c_1 f_1(x)+\cdots+c_m f_m(x)=0$.  If $y=x^3$, so that $y$ and $x^3$ are two different names of the same function, then the linear dependence of the pair $(y,x^3)$ could be considered to be expressed by that equality.  Or one could similarly say the pair $(x^3,x^3)$ is linearly dependent.  However, I wouldn't expect that necessarily to be how the written equality $y=x^3$ will be understood by someone who reads only that and nothing more.
Notice that in this context it matters that the things to which linear dependence or linear independence are attributed are tuples rather than sets or multisets.
