How do you explicitly show that e^(mx) and e^(nx) are independent when m is not equal to n?

I get this intuitively. e^(nx) cannot be generated by multiplying e^(mx) by a constant.

Question

How do you explicitly show that e^(mx) and e^(nx) are independent when m is not equal to n?

I get this intuitively. e^(nx) cannot be generated by multiplying e^(mx) by a constant.

anonymous · Answer

I am not good at English. I want to ask whether 'independent' means not equal?

amistre64 · Answer

can you do a wronskian to settle it?

anonymous · Answer

Yes you can do a Wronskian. I think there is a way to show independence with a matrix though. 

If two function f and g are independent and a, b are real numbers, a*f + b*g = 0 iff a = b = 0.

amistre64 · Answer

$$W=\begin{vmatrix} e^{mx}&e^{nx}\me^{mx}&ne^{nx} \end{vmatrix}=ne^{(m+n)x}-me^{(m+n)x}\ne 0;\ independant$$

amistre64 · Answer

when m is equal to n we get zero

amistre64 · Answer

spose n = a, and m = a

$$ae^{2ax}-ae^{2ax}=e^{2ax}(a-a)\to e^{2ax}(0)=0$$

anonymous · Answer

I see it. thanks. By the way are you typing in LaTex?

amistre64 · Answer

i hope i am :)

turingtest · Answer

nice one amistre :)

amistre64 · Answer

I was trying to stiffle my Laplace of: and therefore the results or clearly obvious that any further proofing would be a waste of my time: kinda attitude ;)