Are functions dependent

Question

How can I show if functions are dependent.

In my case the functions are:

$f_1 = \cos(x)-2x$

$f_2 = x^2\sin(x)$

Is the group $\{f_1,f_2\}$ is lineary independent over R?

Independent on which field? $\mathbb R$? O r on an intervals? — Mikasa, Dec 17 '13 at 13:47

score 3 · Accepted Answer · answered Dec 17 '13 at 14:16

3

A shortcut hint:

If two functions are linearly dependent, then one of them is simply a constant multiple of another.

In fact, if $$cf_2=f_1$$ which means that they are dependent, then $$(-1)f_1+cf_2=0$$ wherein $c\neq 0$ is some constant.

answered Dec 17 '13 at 14:16

Mikasa

67,374

Can you please show me how to solve it once? so I would know how to solve it for next questions? – user844541 Dec 17 '13 at 14:59
@user844541: Practically, the way Mhenni suggested can be extended for more than $2$ functions, but it needs you to know the derivatives. When we have $2$ functions, just examine if one of them is a non-zero multiple of the another or not. – Mikasa Dec 17 '13 at 15:03
it's obvious that they are not dependent, but how can I prove that? – user844541 Dec 17 '13 at 15:07
@user844541: Assume they are L.D., so we have $f_1=cf_2$. Now put $x=\pi$ to find that $c$. Also put $x=\pi/2$. Will the latter resulted $c$ be the same as you found before? – Mikasa Dec 17 '13 at 15:13
but if x=π then $cf2=0$ and you cannot indicate anyhing about c. the equation is just not true for x=π – user844541 Dec 17 '13 at 17:02
@user844541: Oh. I think, I made a mistake in suggesting you to take $x=\pi$. However, you could choose another value fro $x$ like $\pi/4$. – Mikasa Dec 17 '13 at 18:29
@B.S. can't I just use that and show contradiction? (since 0!=1-2π) – user844541 Dec 17 '13 at 19:38
$\ddot\smile +1$ – amWhy Dec 17 '13 at 23:45

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

1

A related problem. Here is how you advance, make the linear combination of the two functions

$$ c_1f_1(x) + c_2f_2(x)=0. $$

Now, we need another equation in $c_1, c_2$. By differentiating the equation with respect to $x$, we get a second equation

$$ c_1f'_1(x) + c_2f'_2(x)=0. $$

We need to solve the two equations for $c_1$ and $c_2$. The system will have a unique solution $c_1=c_2=0$ if The determinant of the matrix of coefficients $A$ of $c_1$ and $c_2$ does not equal $0$. Now, try to find the determinant and see what you get.

Note:

$$|A| = f_1(x)f'_2(x)-f_2(x)f'_1(x).$$

edited Apr 13 '17 at 12:21

Community

1

answered Dec 17 '13 at 13:58

Mhenni Benghorbal

47,431

Is there a way to do that without differentiating the equation? – user844541 Dec 17 '13 at 14:12
@user844541: This method uses techniques from linear algebra which is easy to use and you can handle any finite number of functions and test them. It is a general technique. – Mhenni Benghorbal Dec 17 '13 at 14:19
1

Hello, dear Dr. Mhenni. :+) – Mikasa Dec 18 '13 at 05:49
1

@B.S.: Hello. Thanks for the comment. – Mhenni Benghorbal Dec 18 '13 at 11:10

score 1 · Answer 3 · answered Dec 17 '13 at 14:19

Here's a solution which does not involve differentiation. Assume that $a_1,a_2 \in \mathbb{R}$ satisfy $a_1 f_1(x) + a_2 f_2(x) = 0$ for all $x$ in the domain of the functions, which I'll just assume is $\mathbb{R}$ as well.

Then to show independence, we want to show that $a_1 = a_2 = 0$. One way to go about doing that in this case is by just plugging in values of $x$ for which the functions are easy to evaluate. For instance, at $x = 0$, the equation simplifies to $a_1 = a_1 \cdot 1 + a_2 \cdot 0 = 0$. Similarly, you can cook up a value of $x$ which allows you to deduce that $a_2 = 0$.

Are functions dependent

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