Linear independence

  • Thread starter evagelos
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  • #1
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CAN somebody, please write down a formula defining the linear independence of the following functions??

{[tex] e^x,e^{2x}[/tex]}
 

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  • #2
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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
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The definition says: if [tex]a,b[/tex] are constants and [tex]a e^x + b e^{2x} = 0[/tex] for all [tex]x[/tex], then [tex]a = b = 0[/tex].
 
  • #4
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The definition says: if [tex]a,b[/tex] are constants and [tex]a e^x + b e^{2x} = 0[/tex] for all [tex]x[/tex], then [tex]a = b = 0[/tex].
Would you say that your definition is equivalent to:

for all a,b,x and [tex]ae^x + be^{2x}=0[/tex] ,then a=b=0 ?

or in a more combact form:

for all a,b,x [tex][ae^x + be^{2x} =0\Longrightarrow ( a=b=0)][/tex]
 
  • #5
HallsofIvy
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Would you say that your definition is equivalent to:

for all a,b,x and [tex]ae^x + be^{2x}=0[/tex] ,then a=b=0 ?

or in a more combact form:

for all a,b,x [tex][ae^x + be^{2x} =0\Longrightarrow ( a=b=0)][/tex]
"For all a, b" makes no sense if "a= b= 0".

Just "if, for all x, [tex]ae^x + be^{2x}=0[/tex] ,then a=b=0" or
"(for all x [tex]ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)[/tex]"
 
  • #6
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"For all a, b" makes no sense if "a= b= 0".

Just "if, for all x, [tex]ae^x + be^{2x}=0[/tex] ,then a=b=0" or
"(for all x [tex]ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)[/tex]"
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 [tex] ae^x + be^{2x} =0[/tex] ,then you can prove that the only value a and b can take is zero??
 
  • #7
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[tex](\text{for all } x,\; ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)[/tex]
Incomplete. How about this...

[tex]\text{for all } a,\; \text{for all } b\; [(\text{for all } x,\; ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)][/tex]
 
  • #8
HallsofIvy
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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 [tex] ae^x + be^{2x} =0[/tex] ,then you can prove that the only value a and b can take is zero??
Given that [itex]ae^x+ be^{2x}= 0[/itex] for all x, then, taking x= 0, [itex]a+ b= 0[/itex]. Since [itex]ae^x+ be^{2x}= 0[/itex] for all x, differentiating with respect to x, [itex]ae^x+ 2be^{2x}= 0[/itex] for all x and, setting x= 0 again, [itex]a+ 2b= 0[/itex]. 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?

Incomplete. How about this...

[tex]\text{for all } a,\; \text{for all } b\; [(\text{for all } x,\; ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)][/tex]
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.
 

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