# Homework Help: How to show functions are linearly dependent?

1. Oct 25, 2008

### mikehibbert

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

Show that the set of functions:

x^2
3x+2
x-1
2x+5

are linearly dependent.

2. Relevant equations

-

3. The attempt at a solution

I know that you have to show that you can put constants in front of each equation (that aren't all zero) such that:

c1y1 + c2y2 + c3y3 + c4y4 = 0

i.e. c1(x^2) + c2(3x+2) + c3(x-1) + c4(2x-5) =0

But I have no idea how to do this?

2. Oct 25, 2008

### jacobrhcp

work out c1(x^2) + c2(3x+2) + c3(x-1) + c4(2x-5) =0

by grouping all the same factors of x, ie:

c1x^2 = 0
(3 c2 + c3 + 2 c4)x = 0
... = 0

and figure out what set of constants makes all of these equations true, can you do that?

3. Oct 25, 2008

### mikehibbert

i'm struggling i must admit :S

this would give me:

(c1)x^2 = 0
(3c2 + c3 + 2c4)x = 0
(2c2 + c3 + 5c4) =0

yes?

4. Oct 25, 2008

### Staff: Mentor

Yes. From above, it's clear that c1 = 0, but since you have two more equations in three unknowns (c2, c3, c4) it's pretty likely you're going to get a whole lot of nonzero solutions for these constants.

5. Oct 25, 2008

### mikehibbert

but do you think i have to find the values of the constants?

surely it's impossible? because I have three unknowns - I need three equations?

6. Oct 25, 2008

### HallsofIvy

Therefore what? If the coefficients are nopt all 0, then the functions are dependent!

Last edited by a moderator: Oct 25, 2008
7. Oct 28, 2008

### mikehibbert

I used a wronskian in the end!

8. Oct 28, 2008

### Staff: Mentor

That's like going after a housefly with a bulldozer! It would have been much simpler to just solve this system algebraically:
(c1)x^2 = 0
(3c2 + c3 + 2c4)x = 0
(2c2 + c3 + 5c4) =0

9. Oct 28, 2008

### mikehibbert

but i still cant see how that proves it?

and i needed the marks, so overkill it was :P

10. Oct 28, 2008

### Staff: Mentor

The functions x^2, 3x+2, x-1, and 2x+5 are linearly dependent iff the equation c1x^2 + c2 (3x+2) + c3 (x-1) + c4(2x+5) = 0 has a nontrivial solution. I.e., at least one of the constants ci is nonzero.

You were well on your way to establishing this with this set of equations:
(c1)x^2 = 0
(3c2 + c3 + 2c4)x = 0
(2c2 + c3 + 5c4) =0

I guarantee you, if you can't find a solution to this system (and there are lots of them), you will have a difficult time of it, and knowledge of how to apply the Wronskian will be of little help to you. Guaranteed.
Mark