# [SOLVED] linear independence exponential functions

• May 25th 2009, 02:40 PM
rick81
[SOLVED] linear independence exponential functions
Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.
• May 25th 2009, 02:43 PM
Isomorphism
Quote:

Originally Posted by rick81
Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.

Do you know the polynomials $\displaystyle y,y^2$ and $\displaystyle y^3$ are linearly independent? Hmm.. why am I talking about that? (Itwasntme)
• May 25th 2009, 06:18 PM
ThePerfectHacker
Quote:

Originally Posted by rick81
Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.

Say that $\displaystyle ae^x + be^{2x}+ce^{3x} = 0$ for all $\displaystyle x\in \in \mathbb{R}$.
Then taking the derivative, $\displaystyle ae^x + 2be^{2x} + 3ce^{3x} = 0$.
Take the derivative again, $\displaystyle ae^x + 4be^{2x}+9ce^{3x}=0$.

For any value of $\displaystyle x$ this gives us a $\displaystyle 3\times 3$ system of equation.
The determinant of this system (for this value of $\displaystyle x$) is:
$\displaystyle \left| \begin{array}{ccc} e^x & e^{2x} & e^{3x} \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x} \end{array} \right|$

Show this determinat is not the zero function. Therefore, it would imply that for some $\displaystyle x$ we have a homogenous system with a non-vanishing determinant and hence the trivial solution $\displaystyle a=b=c=0$ is the only such solution.
• May 26th 2009, 01:03 AM
Showcase_22
that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

$\displaystyle a e^x+be^{2x}+ce^{3x}=0$

$\displaystyle e^x(a+be^x+ce^{2x})=0$

$\displaystyle a+be^x+ce^{2x}=0$

$\displaystyle a=-be^x-ce^{2x}$

$\displaystyle a+0e^x+0e^{2x}=-be^x-ce^{2x}$

Comparing coefficients on both sides gives $\displaystyle a=b=c=0$.

?

P.S: 600th post! (Party)
• May 26th 2009, 06:03 AM
HallsofIvy
Quote:

Originally Posted by Showcase_22
that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

$\displaystyle a e^x+be^{2x}+ce^{3x}=0$

$\displaystyle e^x(a+be^x+ce^{2x})=0$

$\displaystyle a+be^x+ce^{2x}=0$

$\displaystyle a=-be^x-ce^{2x}$

$\displaystyle a+0e^x+0e^{2x}=-be^x-ce^{2x}$

Comparing coefficients on both sides gives $\displaystyle a=b=c=0$.

"Comparing coefficients" like that only works if the functions are independent. You are assuming what you want to prove!

Quote:

?

P.S: 600th post! (Party)
• Jun 1st 2009, 11:20 PM
xalk
Quote:

Originally Posted by Showcase_22
that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

$\displaystyle a e^x+be^{2x}+ce^{3x}=0$

$\displaystyle e^x(a+be^x+ce^{2x})=0$

$\displaystyle a+be^x+ce^{2x}=0$

$\displaystyle a=-be^x-ce^{2x}$

$\displaystyle a+0e^x+0e^{2x}=-be^x-ce^{2x}$

Comparing coefficients on both sides gives $\displaystyle a=b=c=0$.

?

P.S: 600th post! (Party)

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

So you proved :

$\displaystyle a +be^x + ce^{2x} = 0$

Now differentiate both sides and you get:

$\displaystyle be^x + 2ce^{2x}=0\Longrightarrow e^x( b + 2ce^x) =0\Longrightarrow b + 2ce^x = 0$

Differentiate again both sides and you get:

$\displaystyle 2ce^x =0\Longrightarrow c=0$

From the equation :$\displaystyle b + 2ce^x=0$ you get b=0

AND from the equation :$\displaystyle a + be^x + ce^{2x} =0$ you get a=0
• Jun 2nd 2009, 11:52 AM
HallsofIvy
You could, of course, do this without differentiating at all (although that is easier!).

You want to find $\displaystyle ae^x+ be^{2x}+ ce^{3x}= 0$. Take x= 0 so you have $\displaystyle a+ b+ c= 0$. Take x= 1 so you have $\displaystyle ae+ be^2+ ce^3= 0$. Take x= 2 so you have $\displaystyle ae^2+ be^4+ ce^6= 0$. 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.