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

  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. dextercioby

    dextercioby 12,324
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    Yes,they are independent.What are u trying to do...?Post the text of the problem...

    Daniel.
     
  4. I'm trying to prove that they are whether dependent or independent. how can I show iy?
     
  5. dextercioby

    dextercioby 12,324
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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. 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. arildno

    arildno 12,015
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    Cecile: What is the DEFINITION of linear independence?
     
  8. dextercioby

    dextercioby 12,324
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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. 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. arildno

    arildno 12,015
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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. dextercioby

    dextercioby 12,324
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    Or you could just compute the wronskian...:uhh:

    Daniel.
     
  12. thank you daniel:)
     
  13. 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. 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. matt grime

    matt grime 9,395
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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. HallsofIvy

    HallsofIvy 40,946
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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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