Does adding a constant to a function makes it linearly independent of each other?

Question

Say, we add $c$ to the function $f(x) = cos(x)$. The new function is $g(x) = cos(x) + c$.

Now are $f$ and $g$ linearly independent of each other?

Using the definition of linear dependence,

$af(x)+bg(x)=0$

it seems that they are linearly independent. Also substituting these in a differential equation, like,

$dy/dx+y=0$

shows that $f(x)$ is a solution but $g(x)$ isn't.

But both of them look very similar in a graph. More similar than multiplying them with a constant (which would have made them dependent).

Am I missing something here?

Also, does two linearly independent functions define some sort of space which span some sort of function dimensions.

I understand linearly independence in vectors but I don't get function equivalent of it.

Answer 1

It might help to treat these functions as vectors:

$v = f(x), w = c$.

Can you find out in which cases $v$ and $v+w$ are linearly dependent and in which they aren't, for completely general vectors $v,w$?
(Hint: consider the case $v = \alpha w$ for some scalar $\alpha$ first)

Furthermore, scalar multiplication will not make things independent. If $\alpha$ is a scalar, then $v$ and $\alpha v$ are always linearly dependent, that is still true if they represent functions.

Or did you want to discuss $f(x)$ and $f(\alpha x)$?

Answer 2

Yes, the functions you describe are linearly independent unless $c=0$.

In the same way, adding a constant (vector) to a vector can make it linearly independent. E.g., $\begin{pmatrix} 1 \\0 \end{pmatrix} + \begin{pmatrix} 0 \\1 \end{pmatrix} =\begin{pmatrix} 1\\ 1\end{pmatrix}$ is linearly independent of $\begin{pmatrix} 1 \\0 \end{pmatrix} $. However, multiplying a vector (or function) by a constant will always yield something that is linearly dependent.

Yes, we can talk about the span of linearly independent functions. In your case, the span of $\cos(x)$ and $\cos(x)+c$ are all functions of the form $\alpha \cos(x) + \beta(\cos(x)+c) = (\alpha+\beta)\cos(x)+\beta c$ for some $\alpha,\beta$. You can check that these are exactly the functions of the form $A \cos (x)+B$ for any values of $A$ and $B$. This vector space has dimension two, since it is the span of two linearly independent vectors.

Answer 3

The condition for linear dependency,

$$\lambda f(x)+\mu(f(x)+c)=0$$

can be written

$$\nu f(x)+\mu c=0.$$

Hence, unless $f(x)$ is a constant function, it is independent of $f(x)+c$.