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

  1. Apr 4, 2005 #1
    [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...
     
  2. jcsd
  3. Apr 4, 2005 #2

    dextercioby

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

    Daniel.
     
  4. Apr 4, 2005 #3
    I'm trying to prove that they are whether dependent or independent. how can I show iy?
     
  5. Apr 4, 2005 #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.
     
  6. Apr 4, 2005 #5
    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
     
  7. Apr 4, 2005 #6

    arildno

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    Cecile: What is the DEFINITION of linear independence?
     
  8. Apr 4, 2005 #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.
     
  9. Apr 4, 2005 #8
    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? :(
     
  10. Apr 4, 2005 #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.
     
  11. Apr 4, 2005 #10

    dextercioby

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    Or you could just compute the wronskian...:uhh:

    Daniel.
     
  12. Apr 4, 2005 #11
    thank you daniel:)
     
  13. Apr 4, 2005 #12
    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?
     
  14. Apr 4, 2005 #13
    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.
     
    Last edited: Apr 4, 2005
  15. Apr 4, 2005 #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.
     
  16. Apr 4, 2005 #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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