# Math Help - [SOLVED] linear independence exponential functions

1. ## [SOLVED] linear independence exponential functions

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

2. Originally Posted by rick81
Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.
Do you know the polynomials $y,y^2$ and $y^3$ are linearly independent? Hmm.. why am I talking about that?

3. Originally Posted by rick81
Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.
Say that $ae^x + be^{2x}+ce^{3x} = 0$ for all $x\in \in \mathbb{R}$.
Then taking the derivative, $ae^x + 2be^{2x} + 3ce^{3x} = 0$.
Take the derivative again, $ae^x + 4be^{2x}+9ce^{3x}=0$.

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

Show this determinat is not the zero function. Therefore, it would imply that for some $x$ we have a homogenous system with a non-vanishing determinant and hence the trivial solution $a=b=c=0$ is the only such solution.

4. that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

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

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

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

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

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

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

?

P.S: 600th post!

5. 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:

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

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

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

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

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

Comparing coefficients on both sides gives $a=b=c=0$.
"Comparing coefficients" like that only works if the functions are independent. You are assuming what you want to prove!

?

P.S: 600th post!

6. 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:

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

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

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

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

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

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

?

P.S: 600th post!

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

So you proved :

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

Now differentiate both sides and you get:

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

Differentiate again both sides and you get:

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

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

AND from the equation : $a + be^x + ce^{2x} =0$ you get a=0

7. You could, of course, do this without differentiating at all (although that is easier!).

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