Using the Wronskian for linear independence/dependence

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Homework Help Overview

The discussion revolves around determining the linear independence or dependence of the functions f1 = 0, f2 = x, and f3 = e^x using the Wronskian. Participants explore the implications of a zero Wronskian and the definitions of linear independence and dependence.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the evaluation of the Wronskian and its implications for linear dependence. Some question the validity of using zero as a constant multiple in linear combinations. Others suggest reconsidering the definitions of linear independence and dependence.

Discussion Status

The discussion is active, with participants exploring different interpretations of the Wronskian's results. There is a recognition that a zero Wronskian does not definitively conclude linear dependence, prompting further inquiry into the definitions and relationships among the functions.

Contextual Notes

Participants note the challenge of demonstrating linear independence without relying on the Wronskian, leading to a deeper examination of linear combinations and scalar coefficients.

bcjochim07
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Homework Statement



f1 = 0 , f2 = x , f3 = e^x

I am supposed to find out if these are linearly independent or dependent. Just by looking at it, I can't see a way to write one of the functions as a combination of the other two with constant multiples, so to make sure that it is linearly independent, I used the Wronskian

f1' = 0 f2' =1 f3'= e^x
f1" = 0 f2"=0 f3" = e^x

So I evaluated the determinant of

0 x e^x
0 1 e^x
0 0 e^x

And that equals zero, which would mean to me that the functions are linearly dependent. I'm not sure how that could be. Any thoughts?

Homework Equations





The Attempt at a Solution

 
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bcjochim07 said:

Homework Statement



f1 = 0 , f2 = x , f3 = e^x

I am supposed to find out if these are linearly independent or dependent. Just by looking at it, I can't see a way to write one of the functions as a combination of the other two with constant multiples
Then you're thinking too narrowly. I bet you've forgotten about the simplest linear combination of two functions.
 
Well... I know that I can't use zero as a constant multiple I think I am overlooking something really simple, but I still can't see how to combine them.
 
bcjochim07 said:
Well... I know that I can't use zero as a constant multiple
Why not?
 
bcjochim07 said:
And that equals zero, which would mean to me that the functions are linearly dependent.

If the Wronskian is zero for all x, then the only thing you can conclude from that, is that the functions may or may not be linearly dependent .

Just because linear dependence implies that the Wronskian is zero, does not mean that the Wronskian being zero implies linear dependence.
 
So how would I go about showing that they are linearly independent if I can't use the Wronskian? Do I just say that there is no way to write one function as a linear combination of the other two?
 
Let's go back to the definition of L.I and L.D.

If I can write c1* f1 +c2* f2 +c3* f3 = 0 for some scalar c1 c2 c3, and not all of them are zero, then we found L.D. So, can you think of anything? Here's a hit, two of the scalars are the same number.
 
gabbagabbahey wasn't telling you that they're linearly independent. He merely said that a zero Wronskian is not proof of linear dependence.

(At least, he better not have been telling you they're linearly independent -- because they really are linearly dependent!)
 
bcjochim07 said:
So how would I go about showing that they are linearly independent if I can't use the Wronskian? Do I just say that there is no way to write one function as a linear combination of the other two?

Well, for constants c_1, c_2 and c_3...the only way that c_1 f_1+c_2 f_2 +c_3 f_3= c_2 x+ c_3 e^x=0 is if c_2=c_3=0, but that doesn't mean c_1 has to be zero does it?
 
  • #10
Ok... I understand. I just reread the definition for linear independence in book, and it all makes sense now. Thanks very much.
 

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