Linear dependency of $f(t) = t^3$ and $g(t) = |t|^3$ I do not understand how to determine the linear dependency of the pair of functions $f(t) = t^3$ and $g(t) = |t|^3$. I know that one method is to check the wronskian of the functions. However, in this case it is equal to $0$ for $t \in (-\infty, 0) \cup (0, +\infty) \equiv \mathbb{R} \setminus \{ 0 \}$. (I am unsure if one can consider the case where $t=0$ separately, in which case the wronskian is also $0$, otherwise it is undefined) So this cannot assert whether the functions are dependent or independent. One must therefore check the following equation:
$$
c_1 f + c_2 g = 0 \\
c_1 \cdot t^3 + c_2 \cdot (\sqrt{t^2})^3 = 0 \\
(c_1 t^3 + c_2 (\sqrt{t^2})^3)(c_1 t^3 - c_2 (\sqrt{t^2})^3) = 0 \\
c_1^2 t^6 - c_2^2 t^6 = 0 \\
t^6 (c_1^2 - c_2^2) = 0 \\
\therefore c_1 = \pm c_2
$$
Meaning that the first line of the above calculations does not imply that $c_1 = c_2 = 0$. So $f(t)$ and $g(t)$ should be dependent. However I know that the "right answer" to this problem (it's for a university course) is 'independent'.
Is my reasoning wrong? I don't see how these two functions could be considered independent.
 A: Suppose $a\,f + b\,g = 0$. That means that $a\,f+b\,g$ is the constant function $0$.
Evaluate at $1$ to get
$a\,f(1)+b\,g(1) = a + b = 0$.
Now evaluate at $-1$ to get
$a\,f(-1)+b\,g(-1) = -a + b = 0$.
Solving the system you get $a=b=0$, so $f$ and $g$ are linearly independent.
A: Let's start...
$$c_1 f + c_2 g = 0 ~\forall t \Rightarrow
c_1  t^3 + c_2 |t|^3 = 0 ~\forall t.
$$
This implies that:
$$\begin{cases}
c_1t^3 + c_2 t^3 = 0 & \forall t \geq 0\\
c_1t^3 - c_2 t^3 = 0 & \forall t < 0
\end{cases} \Rightarrow 
\begin{cases}
t^3(c_1 + c_2) = 0 & \forall t \geq 0\\
t^3(c_1 - c_2) = 0 & \forall t < 0
\end{cases} .
$$
Since the previous equations must hold for every $t$, then:
$$c_1 = c_2 = 0,$$
and hence $f$ and $g$ are linearly independent.
A: The previous answers are totally valid but in my opinion you miss the key argument. As I said in my comment $f$ is odd and $g$ is even so they must be linearly independant. Indeed, forall $a,b \in \mathbb R$ such that $$af+bg=0$$ you get $$af=-bg$$
which is both even and odd. Note $h$ this function, forall $t$ we have
$$
h(t)=-h(-t)=-h(t)
$$
and thus $h(t) = 0$. Since $f$ and $g$ are not null, $a=b=0$ and you are done.
See that I didn't computed the value of $f$ or $g$ but only used their parity. This argument can be easily generalized.
