# Linear independence

1. Jul 1, 2009

### evagelos

CAN somebody, please write down a formula defining the linear independence of the following functions??

{$$e^x,e^{2x}$$}

2. Jul 1, 2009

### Pinu7

If e^x and e^2x are linearly independent then the equation

ae^x+be^2x=0 must have the only solution a,b=0.

Differentiate to obtain,

ae^x+2be^2x=0

Subtract the first equation from the second to obtain,
be^2x=0
which is true if and only if b=0.
Then it is obvious that a=0.

Therefore, e^x and e^2x are linearly independent.

3. Jul 1, 2009

### g_edgar

The definition says: if $$a,b$$ are constants and $$a e^x + b e^{2x} = 0$$ for all $$x$$, then $$a = b = 0$$.

4. Jul 1, 2009

### evagelos

Would you say that your definition is equivalent to:

for all a,b,x and $$ae^x + be^{2x}=0$$ ,then a=b=0 ?

or in a more combact form:

for all a,b,x $$[ae^x + be^{2x} =0\Longrightarrow ( a=b=0)]$$

5. Jul 1, 2009

### HallsofIvy

Staff Emeritus
"For all a, b" makes no sense if "a= b= 0".

Just "if, for all x, $$ae^x + be^{2x}=0$$ ,then a=b=0" or
"(for all x $$ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)$$"

6. Jul 2, 2009

### evagelos

You mean that a and ,b cannot have any other value apart from zero??

In that case you do not have to prove anything because ae^x + be^2x =0

OR given any a,b,x and if $$ae^x + be^{2x} =0$$ ,then you can prove that the only value a and b can take is zero??

7. Jul 2, 2009

### g_edgar

$$\text{for all } a,\; \text{for all } b\; [(\text{for all } x,\; ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)]$$
Given that $ae^x+ be^{2x}= 0$ for all x, then, taking x= 0, $a+ b= 0$. Since $ae^x+ be^{2x}= 0$ for all x, differentiating with respect to x, $ae^x+ 2be^{2x}= 0$ for all x and, setting x= 0 again, $a+ 2b= 0$. Subtracting the first equation from the second, (a+ 2b)- (a+ b)= b= 0. Putting b= 0 in either equation, a= 0. Is that what you are asking?