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Linear independence of sin (x), cos (x) and 1, proof

  1. Apr 23, 2012 #1
    What would be the best way to show that functions [itex]f(x)=1[/itex], [itex]g(x)=sin(x)[/itex] and [itex]h(x)=cos(x)[/itex] are linearly independent elements of the vector space [itex]\mathbb{R}^{\mathbb{R}}[/itex]?

    I know that the linear independence means that an expression like [itex]\alpha \mathbb{x}_1 + \beta \mathbb{x}_2 + \gamma \mathbb{x}_3 = \mathbb{0}[/itex] is true only for [itex]\alpha = \beta = \gamma = 0[/itex] where [itex]x_1,...,x_3[/itex] are vectors and [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] are scalars of the vector space.

    I think that the proof might look like this:

    [itex]\alpha sin(x)+ \beta cos(x)+ \gamma 1=0[/itex]

    If [itex]x=0[/itex] then [itex]sin(x)=0[/itex]. Therefore, [itex]\beta=0[/itex] and [itex]\gamma=0[/itex], but [itex]\alpha[/itex] might be different than zero, and the above-mentioned expression still equal to zero.
  2. jcsd
  3. Apr 23, 2012 #2
    Your attempt is a good one. So assume that there are [itex]\alpha,\beta,\gamma[/itex] such that

    [tex]\alpha f + \beta g+\gamma h=0[/tex]

    That means that for ALL x must hold that

    [tex]\alpha+\beta\sin(x)+\gamma \cos(x)=0[/tex]

    This holds for all x, so try to pick some good values for x.

    You already tried x=0, this gives us that necessarily


    (and not [itex]\alpha=0,\gamma=0[/itex] as you claimed).

    Now try some other values for x. For example pi or pi/2 ??

    PS excuse me for using other [itex]\alpha,\beta,\gamma[/itex] as in your post.
  4. Apr 23, 2012 #3
    For [itex]x=\pi[/itex], we get [itex]\gamma - \beta = 0[/itex] which means that [itex]\alpha[/itex] can be of any value, and the expression still equal to zero. Then those elements ([itex]f(x)[/itex], [itex]g(x)[/itex] and [itex]h(x)[/itex]) would not be linearly independent according to the definition of linear independence. I think that we need all three scalars to be zero to prove the linear independence: [itex]\alpha =0[/itex], [itex]\beta =0[/itex] and [itex]\gamma = 0[/itex]. In other words, [itex]sin(x)\neq 0[/itex] and [itex]cos(x)\neq 0[/itex].

    For [itex]x=\frac{\pi}{3}[/itex], we get [itex]\frac{\sqrt{3}}{2}\alpha +\frac{1}{2}\beta + \gamma = 0[/itex], which means that [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] must be equal to zero for the expression to be true.
  5. Apr 23, 2012 #4
    Why should this imply that [itex]\alpha,\beta,\gamma[/itex] are all zero?? It doesn't.
  6. Apr 23, 2012 #5
    It does if we want to prove the linear independence (because of the definition itself). I'm worried about the fact that not all [itex]x[/itex] satisfy the conditions [itex]sin(x)\neq 0[/itex], [itex]cos(x)\neq 0[/itex] that allow us to prove it.
  7. Apr 23, 2012 #6


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    Because you want them all equal to 0, you simply declare that
    [tex]\frac{\sqrt{3}}{2}\alpha+ \frac{1}{2}\beta+ \gamma= 0[/tex]?
    Looks like you are assuming what you want to prove.

    What about [itex]\alpha= 0[/itex], [itex]\beta= 2[/itex], [itex]\gamma= -1[/itex]?

    To prove that 1, sin(x), and cos(x) are independent, you want to prove that the only way you can have [itex]\alpha (1)+ \beta(sin(x))+ \gamma(cos(x))= 0[/itex] for all x is to have [itex]\alpha= \beta= \gamma= 0[/itex]. But that is what we want to prove- we cannot assume it.

    Since that is true for all x, it is, in particular, true for x= 0, we must have
    [itex]\alpha+ \gamma= 0[/itex]
    And, for [itex]x= \pi/2[/itex], we must have
    [itex]\alpha+ \beta= 0[/itex]

    Finally, for [itex]x= \pi[/itex], we must have
    [itex]\alpha- \gamma= 0[/itex]

    Solve those three equations for [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex].
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