# Linearly dependent or independent functions?

1. Aug 31, 2007

### kasse

Are the functions y=0 and y=sinh(pi*x) linearly dependent or linearly independent on the intercal x>0?

I'm not sure what I'm supposed to do here, but I try to divide them:

0/sinh(pi*x). This is certainly 0, since sinh (pi*x) is positive when x>0. Since 0 is a constant, the functions must be linearly dependent.

Am I right?

2. Aug 31, 2007

### wbclark

Any set that contains the 0 vector is linearly dependent.

In this case, the zero function is the zero vector of the space of functions defined on [o,infinity).

This is because any vector written as a linear combination of linearly independent vectors should have a unique linear combination... however, with the 0 vector included, you can come up with as many linear combinations as you want.

Hope this helps

3. Aug 31, 2007

### kasse

Thanks. I haven't learnt that much about matrixes yet. But the way I reasoned is plausible as well?

4. Aug 31, 2007

### wbclark

It really depends on what type of class this is for.

Is this a linear algebra class?

5. Aug 31, 2007

### kasse

Yes. This chapter is called "Second order linear ODEs".

Another problem involves the functions ln x and ln(x^2). Then I wrote ln(x^2) as 2ln x, and since ln x/2ln x = 1/2, the functions are lineary dependent.

Last edited: Aug 31, 2007
6. Aug 31, 2007

### wbclark

Ahhh, ok.

I would probably just point out that the functions are linearly dependent because one can be written as a linear combination of the other.

Let f(x) = 0
Let y(x) = sin(pi*x)

f(x) = 0 = 0 * sin(pi*x) = 0 * g(x)

Since f(x) is written here as a constant times g(x), you have linearly dependent functions.

7. Sep 1, 2007

### HallsofIvy

Staff Emeritus
I was a little puzzled why you divided them. A set of vectors is "independent" if and only if the only linear combination,
$$\alpha_1 v_1+ \alpha_2 v_2+ \cdot\cdot\cdot+ \alpha_n v_n= 0$$
must have all the [/alpha]s equal to 0.
Of course, for just two vectors (or functions) that is
$$\alpha_1 f+ \alpha_2 g= 0$$
must have $\alpha_1= \alpha_2= 0$. If that's not true, if the functions are dependent, then
$$\frac{f}{g}= -\frac{\alpha_1}{\alpha_2}$$
exists and is non 0. So your method is correct.