0
$\begingroup$

I have 4 functions.

$f_{1} = \cos(x)$ $f_{2} = \sin(x)$ $f_{3} = e^x\cos(x)$ $f_{4} = e^x\sin(x)$

Is $\{f_{1}, f_{2}, f_{3}, f_{4}\}$ linearly independent in $C(\mathbb{R})$? My inital conclusion is no because the linear combination of these functions $af_{1} + bf_{2} + cf_{3} + df_{4} = 0$ for $a=b=c=d=1\neq 0$ namely for when $x = 3\pi/4$. However, I'm not sure if my reasoning is correct. Any help in explaining whether it is linearly independent or dependent would be helpful. Thank you.

$\endgroup$

2 Answers 2

3
$\begingroup$

Hint: Linear dependence means we have $af_1(x)+bf_2(x)+cf_3(x)+df_4 (x)=0$ for every $x$ (with not all coefficients $0$), not just for one particular value of $x$.

Put $x=0, x=\pi, x=\pi /2$ and $x =\pi /3$. (Do you know the values of $\sin x$ and $\cos x$ for these values of $x$?). You will get $4$ equations for $a,b,c,d$. Try to show from these equations that $a=b=c=d=0$.

$\endgroup$
1
$\begingroup$

To prove that these functions aren't independent you would need to prove that there exists a nontrivial linear combination that forms the zero vector in that space. In your case, this means $af_1 + bf_2 + cf_3 + df_4 = f_0$ where $f_0$ is the constant function defined as $f_0(x) = 0$ for all $x\in \mathbb{R}$.

It is not enough to see that the resulting function is zero at a specific point such as $x=3\pi/4$. It must be null everywhere, since this is the property that defines the zero function in $\mathit{C}(\mathbb{R})$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .