Are functions dependent How can I show if functions are dependent.
In my case the functions are:
$f_1 = \cos(x)-2x$
$f_2 = x^2\sin(x)$
Is the group $\{f_1,f_2\}$ is lineary independent over R?
 A: A shortcut hint:

If two functions are linearly dependent, then one of them is simply a constant multiple of another.

In fact, if $$cf_2=f_1$$ which means that they are dependent, then $$(-1)f_1+cf_2=0$$ wherein $c\neq 0$ is some constant.
A: A related problem. Here is how you advance, make the linear combination of the two functions
$$ c_1f_1(x) + c_2f_2(x)=0. $$
Now, we need another equation in $c_1, c_2$. By differentiating the equation with respect to $x$, we get a second equation
$$ c_1f'_1(x) + c_2f'_2(x)=0. $$
We need to solve the two equations for $c_1$ and $c_2$. The system will have a unique solution $c_1=c_2=0$ if The determinant of the matrix of coefficients $A$ of $c_1$ and $c_2$ does not equal $0$. Now, try to find the determinant and see what you get.
Note:

$$|A| = f_1(x)f'_2(x)-f_2(x)f'_1(x).$$

A: Here's a solution which does not involve differentiation. Assume that $a_1,a_2 \in \mathbb{R}$ satisfy $a_1 f_1(x) + a_2 f_2(x) = 0$ for all $x$ in the domain of the functions, which I'll just assume is $\mathbb{R}$ as well.
Then to show independence, we want to show that $a_1 = a_2 = 0$. One way to go about doing that in this case is by just plugging in values of $x$ for which the functions are easy to evaluate. For instance, at $x = 0$, the equation simplifies to $a_1 = a_1 \cdot 1 + a_2 \cdot 0 = 0$. Similarly, you can cook up a value of $x$ which allows you to deduce that $a_2 = 0$.
