Linear Independence: Determining w/ Wronskian Matrix

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Homework Help Overview

The discussion revolves around determining the linear independence of the functions e^x, sin(x), and cos(x) using the Wronskian determinant. Participants are exploring the implications of the Wronskian's properties in relation to linear dependence.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the Wronskian determinant and concludes that the functions are linearly independent based on the result. Other participants question the necessity of the Wronskian being non-zero at all points versus just at some points.

Discussion Status

The discussion is active, with participants clarifying the conditions under which the Wronskian indicates linear independence. Some guidance has been provided regarding the implications of the determinant being non-zero at specific points.

Contextual Notes

Participants are navigating assumptions about the behavior of the Wronskian and its relationship to linear dependence, particularly in the context of the functions being analyzed.

dashkin111
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Homework Statement


Using the wronskian (determinant basically), determine if e^x, sin(x), cos(x) are linearly independent



Homework Equations


I used this:
[tex] | e^{x} sin(x) \:cos(x)|[/tex]
[tex] |e^{x} cos(x) -sin(x)|[/tex]
[tex]|e^{x} -sin(x) -cos(x)|[/tex]

But pretend that's just a 3x3 matrix and you take the determinant of it



The Attempt at a Solution



After finding the determinant I get -2e^x which is never 0 so they're linearly independent. Am I right in this?
 
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Yes, but you know the wronskian only has to be nonvanishing someplace for the functions to be linearly independent, right?
 
Dick said:
Yes, but you know the wronskian only has to be nonvanishing someplace for the functions to be linearly independent, right?

Oh wow, I thought it was at all points. If possible, would you mind telling me why that is?
 
Sure. If f1(x), f2(x) and f3(x) are linearly dependent, then there are nonzero constants such that c1*f1(x)+c2*f2(x)+c3*f3(x) is identically zero over some interval. So a linear combination of columns in your matrix is zero. This tells you det=0 over the interval. So if det is non-zero anywhere, you know they aren't linearly dependent.
 
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