[sin(t) , cos(t), e^t] are these linearly dependent?

  • Thread starter Cecile
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[sin(t) , cos(t), e^t] are these linearly dependent?


can someone solve this q?
I write

Asint+BcosT+C(e^t) = 0

but then i cannot proceed...
 

Answers and Replies

  • #2
dextercioby
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Yes,they are independent.What are u trying to do...?Post the text of the problem...

Daniel.
 
  • #3
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I'm trying to prove that they are whether dependent or independent. how can I show iy?
 
  • #4
dextercioby
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Are u trying to prove that they can form a basis in some vector space...?Define the vector space.

Daniel.
 
  • #5
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No it just asks whether they are dep or in dependent.
for ex if it would have given me sets like v=(0,2,3) , u=(1,2,3,) & z=(1,1,0) then i could write
Av+Bu+Cz=0 and I could show that they are indep of not just by assuming one of the coef. is not zero. but i cannot show this with cos sin & e
 
  • #6
arildno
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Cecile: What is the DEFINITION of linear independence?
 
  • #7
dextercioby
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Here's a nice way to do it.Use differential equations.Consider the homogenous,linear,constant coefficient ODE

[tex] \frac{d^{4}y(x)}{dx^{4}}-y(x)=0 [/tex]

Daniel.
 
  • #8
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if we dont have nontrivial solution for the combination of n vectors then these n vectors are said to be linearly dependent.

I know the DEFINITION just i cannot show that rule for sin cos & e
?!? am I not clear yet? :(
 
  • #9
arildno
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Remember the vector is the WHOLE function here; that is:
If you can show that the following equation,

[tex]a_{1}\sin(t)+a_{2}\cos(t)+a_{3}e^{t}=0[/tex]

in order to be valid (that is, holds) for ALL values of "t" implies that [tex]a_{1}=a_{2}=a_{3}=0[/tex]
then you have concluded that the three functions are linearly independent.
 
  • #10
dextercioby
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Or you could just compute the wronskian...:uhh:

Daniel.
 
  • #11
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thank you daniel:)
 
  • #12
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less machinery

Can I suggest an idea that doesn't involve the machinery of the Wronskian.
First consider the periodicity of the three functions. What can you conclude about the coefficient of [tex]e^x[/tex].
Now consider a root of cos. What does this tell you about the coefficient of sin. Now what must the coefficient of cos be?
 
  • #13
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Or a (very tiny little) bit cleaner, note that we just need

[tex]A\sin{x} + B\cos{x} + Ce^x \equiv 0 \Longrightarrow A=B=C=0[/tex]

setting [itex]x=0[/itex] immediately gives [itex]B+C = 0 \Longrightarrow B=-C[/itex]. But as you noted the limiting behaviour of [itex]e^x[/itex] at infinity implies [itex]C=0[/itex] so [itex]B=-C=0[/itex]. But then [itex]A\sin{x}\equiv 0[/itex] obviously implies [itex]A=0[/itex] so we're done.
 
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  • #14
matt grime
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Yes, thank goodness someone came up with the sensible and obvious approach. No nonn-trivial combination can be the zero *function*, and that can be gotten just from putting some values of x in.
 
  • #15
HallsofIvy
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Saying that sin(x), cos(x), and ex are independent means that

In order for C1sin(x)+ C2cos(x)+ C3ex[/sub]= 0 for all x, we must have C1= C2= C3.

Take 3 different values for x:

x= 0 is especially easy: if x= 0, C1sin(x)+ C2cos(x)+ C3ex[/sub]= 0 becomes C2+ C3= 0.
If [tex]x= \frac{\pi}{2}[/b], [tex]C_1+ e^{\frac{\pi}{2}}C_3= 0[/tex].

Okay, [tex]C_2= -C_3[/tex] and [tex]C_1= -e^{\frac{\pi}{2}}C_3[/tex].
Now put those into the original equation and take x to be some third number. Solve that for C3. If C3= 0 then so must C1= 0 and
C2= 0 and the functions are independent.

Yes, using the Wronskian is simpler.
 

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