Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.

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- May 25th 2009, 03:40 PMrick81[SOLVED] linear independence exponential functions
Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.

- May 25th 2009, 03:43 PMIsomorphism
- May 25th 2009, 07:18 PMThePerfectHacker
Say that for all .

Then taking the derivative, .

Take the derivative again, .

For any value of this gives us a system of equation.

The determinant of this system (for this value of ) is:

Show this determinat is not the zero function. Therefore, it would imply that for some we have a homogenous system with a non-vanishing determinant and hence the trivial solution is the only such solution. - May 26th 2009, 02:03 AMShowcase_22
that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

Comparing coefficients on both sides gives .

?

P.S: 600th post! (Party) - May 26th 2009, 07:03 AMHallsofIvy
- Jun 2nd 2009, 12:20 AMxalk

Showcase ,i will to extent your proof and try to solve the problem without using determinants.

So you proved :

Now differentiate both sides and you get:

Differentiate again both sides and you get:

From the equation : you get b=0

AND from the equation : you get a=0 - Jun 2nd 2009, 12:52 PMHallsofIvy
You could, of course, do this without differentiating at all (although that is easier!).

You want to find . Take x= 0 so you have . Take x= 1 so you have . Take x= 2 so you have . You now have three linear equations to solve for a, b, and c. Do that and show that a= b= c= 0 is the only solution.