linear independence

**evagelos** · Jul1-09, 10:45 AM

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

{ $LaTeX Code: e^x,e^{2x}$ }

**Pinu7** · Jul1-09, 11:05 AM

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.

**g_edgar** · Jul1-09, 11:06 AM

The definition says: if

are constants and $LaTeX Code: a e^x + b e^{2x} = 0$ for all

, then

.

**evagelos** · Jul1-09, 03:22 PM

Originally Posted by g_edgar

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

Would you say that your definition is equivalent to:

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

or in a more combact form:

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

**HallsofIvy** · Jul1-09, 06:11 PM

Originally Posted by evagelos

Would you say that your definition is equivalent to:

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

or in a more combact form:

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

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

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

**evagelos** · Jul2-09, 10:21 AM

Originally Posted by HallsofIvy

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

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

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 $LaTeX Code: ae^x + be^{2x} =0$ ,then you can prove that the only value a and b can take is zero??

**g_edgar** · Jul2-09, 12:13 PM

Originally Posted by HallsofIvy

$LaTeX Code: (\\text{for all } x,\\; ae^x + be^{2x} =0)\\Longrightarrow ( a=b=0)$

Incomplete. How about this...

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

**HallsofIvy** · Jul2-09, 01:39 PM

Originally Posted by 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 $LaTeX Code: ae^x + be^{2x} =0$ ,then you can prove that the only value a and b can take is zero??

Given that $LaTeX Code: ae^x+ be^{2x}= 0$ for all x, then, taking x= 0,

. Since $LaTeX Code: ae^x+ be^{2x}= 0$ for all x, differentiating with respect to x, $LaTeX Code: ae^x+ 2be^{2x}= 0$ for all x and, setting x= 0 again,

. 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?

Originally Posted by g_edgar

Incomplete. How about this...

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

NO!! It makes NO sense say "for all a, a= 0"! "If, for all x, ax= 0, then a= 0." It makes no sense to say "for all a" there.