Linearly Independent of exponenial functions

**Liverpool** · November 30th 2010, 04:55 AM

The problem :

Show that the functions $e^{ax}$ , $e^{bx}$ and $e^{cx}$ are linearly independent if $a \neq b \neq c$

Solution :
I calculated the wronskian, and my final answer is:

$e^{(a+b+c)x} \left( (b-a)c^2+(a-c)b^2+(c-b)a^2 \right)$

Clearly, $e^{(a+b+c)x} \neq 0$

The problem here is with $(b-a)c^2+(a-c)b^2+(c-b)a^2$

Since $a \neq b \neq c$ , the following numbers can't be zero:

b-a
a-c
c-b

Now, I stopped!

How I can prove that the expression $(b-a)c^2+(a-c)b^2+(c-b)a^2$ can't be zero so that wronskian will be non zero and hence the function are linearly independent?

**Ackbeet** · November 30th 2010, 05:01 AM

[EDIT]: Your Wronskian is correct, just not simplified as much as it could be. If you simplify it a bit further, I think your result will pop right out.

**Liverpool** · November 30th 2010, 05:07 AM

I do not see any further simplification :S

**Ackbeet** · November 30th 2010, 05:33 AM

Well, you're trying to set the expression equal to zero, and see under what conditions that can happen, right? So you have a quadratic in $a$ , for example. Try solving

$(b-a)c^{2}+(a-c)b^{2}+(c-b)a^{2}=0$ for $a.$

What do you get?

**Liverpool** · November 30th 2010, 09:41 AM

it will be:

$(c-b)a^2+(b^2-c^2)a+(bc^2-cb^2)=0$

here:

$a'=c-b , b'=b^2-c^2 & c'=bc^2-cb^2$

so:

$a_{1,2}=\dfrac{-b' \pm \sqrt{ (b')^2 - 4a'c' } } { 2a' }$

it will be too long to solve it since i should find $(b')^2 , 4a'c'$ ... etc

Anyone could explain?

**Ackbeet** · November 30th 2010, 09:51 AM

You should carry your computation through. It's not that much algebra. I think you'll find that your answers simplify down a great deal.

**Liverpool** · November 30th 2010, 09:53 AM

This is what I can do.
I do not see any further simplification.

**Ackbeet** · November 30th 2010, 09:56 AM

Just plug in your primed variables. Try things. Note that $a^{2}-b^{2}=(a-b)(a+b).$

The simplification is, for me, about 5 lines' worth. If you're taking differential equations, you should definitely be prepared to do that much algebra.

This forum is about helping people get unstuck. It is not about giving people answers. So plug things in, show your work, and let me know if you get stuck.

**Liverpool** · November 30th 2010, 10:05 AM

xD.
yup yup it will be reduced to:

$a^2-(b+c)a+bc=0$

So the solution for this equation is:

$a_{1,2}=\dfrac{b+c \pm (b-c)}{2}$

so ?

**Ackbeet** · November 30th 2010, 10:08 AM

Keep going. Take the plus sign solution: what is a? Take the negative sign solution: what is a?

**Liverpool** · November 30th 2010, 10:09 AM

The two solution are: b and c.
so?

**Ackbeet** · November 30th 2010, 10:11 AM

So, you could factor the original expression

$(b-a)c^{2}+(a-c)b^{2}+(c-b)a^{2}$

as

$(a-c)(a-b)x,$

where $x$ is something or other. Do you think you could figure out what $x$ is? What do you want it to be?

**Liverpool** · November 30th 2010, 10:14 AM

Well,

$x=\dfrac{(b-a)c^{2}+(a-c)b^{2}+(c-b)a^{2}}{(a-c)(a-b)}$

I do not what I want it to be!

**Ackbeet** · November 30th 2010, 10:18 AM

Consider the symmetry of the original expression. a, b, and c are all treated the same, right? You could probably replace a with b, b with c, and c with a without changing the expression (or maybe you'd get a minus sign - that's not important). But (a-c)(a-b) is not that symmetric. How could you make it equally symmetric? Another way of phrasing it is this: what is missing?

**Liverpool** · November 30th 2010, 10:21 AM

Sorry, but I did not understand "Consider the symmetry of the original expression".

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