Homework Help: Test the set of functions for linear independence in F

1. Feb 19, 2012

Sun God

1. The problem statement, all variables and given/known data

Test the set of {1, ln(2x), ln(x^2)} for linear independence in F, the set of all functions.

If it is linearly dependent, express one of the functions as a linear combination of the others.

2. Relevant equations

N/A

3. The attempt at a solution

I know if [ a(1) + b(ln(2x)) + c(ln(x^2)) = 0 ] implies [ a = b = c = 0 ], then the set is linearly independent. Otherwise, it is linearly dependent.

The book gives an example problem where you can plug in 0 for x and solve for the coefficients, which are then shown to be 0. But that doesn't really work here.

Plugged in x=1: a+b(ln2) + cln(1) =0 implies a + bln2 = 0.

It works neatly in the book example (sinx, cosx) but doesn't really in this case. I can't think of any way to go about doing this...

2. Feb 20, 2012

LCKurtz

Try simplifying your ln functions using properties of logarithms and see if that gives you any ideas.

3. Feb 21, 2012

Sun God

Thank you for the help. Apologies again if I'm missing something that's obvious, but I still cannot find the answer. This is my process:

a(1) + b(ln2x) + c(ln(x^2))

= a(1) + b(ln2 + lnx) + c(2lnx)

= a + bln2 + blnx + 2clnx

= a + b*ln2 + (b+2c)*lnx

After trying to play around with it some more, set up some kind of system of equations, reason it out, etc. I still can't find some way to prove or disprove that a = b = c = 0. Any more suggestions? :\

Last edited: Feb 21, 2012
4. Feb 21, 2012

Dick

That's good. So you've got a+b*ln(2)=0 and b+2c=0. How about putting c=1 and b=(-2). Now what's a?

5. Feb 21, 2012

Sun God

Okay, to follow the above poster:

I have a system of equations of

a + bln2 = 0
b + 2c = 0

I set c = 1 and b = -2

Then I get a = 2ln2

Plug this back into the original equation to get

2ln2 - 2ln2 - 2lnx + 2lnx = 0

Which is always true, therefore the functions ARE linearly independent.

My understanding of some of these concepts is probably shaky, since reading through this I still haven't managed to convince myself that this is true (like for instance, how did the values for b and c come about? guess and check?). But I will go over the concepts several times and see if it sticks, thanks everyone!

6. Feb 21, 2012

Dick

No, no. Finding a nonzero a, b, c shows that they are linearly dependent. Not independent!

7. Feb 22, 2012

Sun God

Ahaha... no wonder it wasn't making sense to me. It's clearer now, much thanks for your patience!

(To clarify for myself and maybe others: it was a matter of finding nonzero coefficients a, b, c and still have the linear combination of the functions equal to zero; if we found such a, b, and c, then the set of functions functions cannot be linearly independent.)

So to finish up the problem, I would show linear dependence by writing one function in terms of the other:

First plug in our values for a, b, c: (2ln2)(1) + (-2)(ln2x) + (ln(x^2)) = 0

which equals:

2ln2 + ln(x^2) = 2ln2x

which we can confirm by playing with ln properties:

2ln2 + 2lnx = 2ln2x
2(ln2 + lnx) = 2ln2x
2ln2x = 2ln2x

Okay. Thanks again.