Show that the functions are linearly independent

Question

Given the following functions:

$f_1(x) = e^x$,

$f_2(x)=sin(x)$,

$f_3(x) =cos(x)$,

$f_4(x) =x$,

$f_5(x) =1$

I have to show that they build a basis of a subspace of a space of all functions from $\Bbb R$ to $\Bbb R$. So I have to show, that:

$k_1f_1(x) + k_2f_2(x) + k_3f_3(x) + k_4f_4(x) + k_5f_5(x) = 0 \Leftarrow \Rightarrow k_1=k_2=k_3=k_4=k_5=0 $

Could you please help me to show this? Which method I should use? I'm not allowed to use Wronskian.

Evalute the functions at various values of $x$. For instance, $x=0$, will give you $k_1 + k_3 + k_5 = 0$. — Prahlad Vaidyanathan, Dec 18 '13 at 09:59
for x = 0 is clear to me. But which further values i should choose for trigon.functions? Should it be values with pi? what to do with $e^{pi}$ then? — Ina, Dec 18 '13 at 10:09

score 1 · Answer 1 · answered Dec 18 '13 at 10:44

You have at least two ways of solving this kind of exercise. You can use purely algebraic methods, that is, you evaluate your linear combination at various points, giving relations between the $k_i$, and, with sufficiently many relations, you will find that the only solution is that all $k_i$ are zero. This can be long and tedious.

But you can also use analytic methods. Suppose that $\sum k_i f_i$ is zero. This means that for all $x$, $\sum k_i f_i(x)$ is zero. When $x$ tends to infinity, this sum is equivalent to $k_1 e^x$, if $k_1$ is not zero. Since the sum is constanly equal to $0$, this proves that $k_1$ has to be $0$. So, in fact, the sum is equivalent to $k_4 x$ (the other functions are bounded) if $k_4$ is not zero. So this proves that $k_4$ is zero.

Derive your relation, and evaluate at $0$, you find $k_2 = 0$. Derive once more, and evaluate at $0$, you find $k_3 = 0$. Hence, $k_5$ is also zero.

Remark : You can also write the Taylor series of your sum, instead of evaluate at various points. If you write at a sufficiently high order, this will show that all $k_i$ are zero. For instance, let $f(x) = \log(1+x)$. Prove that $f$, $f \circ f$ and $f \circ f \circ f$ are linearly independant where they are defined.

score 0 · Answer 2 · answered Dec 18 '13 at 10:02

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Hint check whether Wronskian is non-zero.

answered Dec 18 '13 at 10:02

Myshkin

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not allowed to use Wronskian. – Ina Dec 18 '13 at 10:13
@Lina: But you tagged the question by vector space, so I think you already know the determinant, and of course, the derivatives o elementary functions as well. – Mikasa Dec 18 '13 at 10:33

Show that the functions are linearly independent

2 Answers2