- #1
Luka
- 6
- 0
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.
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.