3
$\begingroup$

Suppose there are three vectors $\sin x,\cos x,\tan x$. Now I'm pretty sure that these lie on the same plane. But can I find any constant coefficients that satisfies $c_1\sin x + c_2\cos x + c_3\tan x =0$? I couldn't. Now please rectify me if there is some way to prove that they are linearly dependent. Or please explain if they aren't linearly dependent how they are lying on the same plane ($xy$ plane) .

$\endgroup$
8
  • 6
    $\begingroup$ The comment you posted on my answer raises an interesting question! All the answers interpreted your question as are the three reals maps $\sin, \cos$ and $\tan$ linearly independent?. Now strictly speaking, it is not the question you asked. But then I have a question for you! In what vector space are you locating the three vectors $\sin x$, $\cos x$ and $\tan x$? I'm convinced you'll get good knowledge answering that question. $\endgroup$ Aug 24, 2018 at 9:29
  • $\begingroup$ These are transcendence functions and they are linearly independent also in any interval [a, b]. $\endgroup$
    – dmtri
    Aug 24, 2018 at 9:31
  • $\begingroup$ @mathcounterexamples.net looks like you have to give me a link to make me know what vector space is and it's properties. And how this disables me to map these functions ? $\endgroup$
    – user187604
    Aug 24, 2018 at 9:38
  • 2
    $\begingroup$ Wikipedia is a good start. When you speak of vectors, those vectors belong to a vector space. The fact that they are linearly independent depends on how you look at them. And in particular to which vector space they belong. $\endgroup$ Aug 24, 2018 at 9:42
  • 1
    $\begingroup$ Note to the asker: On stackexchange websites, it's generally a bad idea to accept an answer within minutes of it being posted. Especially in this case where there's clearly so much confusion surrounding the question: I suspect mathcounteexamples.net's answer is not actually answering the question you intended to ask. $\endgroup$
    – Jack M
    Aug 24, 2018 at 12:28

6 Answers 6

14
$\begingroup$

Beware: strong confusion between vectors and functions!

Three coplanar vectors in $\mathbb{R}^3$ are indeed not linearly independent...

But the example you are giving is not three vectors, these are three functions... This is completely different.

And in that case, the three functions are clearly independent from each other, but these functions are not elements of $\mathbb{R}^3$

EDIT: side note: they are represented in the $xy$ plane, this is not to say they are elements of thus plane...

$\endgroup$
5
  • 1
    $\begingroup$ What if I asked $sinx\hat{i}$ , $cosx\hat{i}$ , and $tanx\hat{i}$ are in the same plane or not . They are now vector fields in fact. $\endgroup$
    – user187604
    Aug 24, 2018 at 9:40
  • $\begingroup$ @user187604 If you take the three vectors $(x_0,\sin x_0), (x_0,\cos x_0), (x_0,\tan x_0)$, then they are linearly dependent... But you take the single vectors given by the initial input point and the image by each function... Not the graph of each function... $\endgroup$
    – Martigan
    Aug 24, 2018 at 9:47
  • 2
    $\begingroup$ This should be an accepted answer. This is the root cause. OP tries to apply a law for vectors to vector functions and gets "incorrect" results. $\endgroup$
    – Ister
    Aug 24, 2018 at 11:07
  • $\begingroup$ There are many vector spaces those functions are vectors in. Like, the space of functions from $\mathbb{R}$ to $\mathbb{R}$. Maybe it would help the OP if you mention that, and how they are not linearly independent in the space in which they are vectors? $\endgroup$
    – Yakk
    Aug 24, 2018 at 13:39
  • $\begingroup$ @Yakk You are right, but I was under the impression it has the potential to make it even more complex. On top of that there are many different vector spaces these functions are vector in... It complexifies further the issue. $\endgroup$
    – Martigan
    Aug 24, 2018 at 14:05
2
$\begingroup$

There are no such constants (unless they're all $0$; of course). In fact,$$x=0\implies c_1\sin(x)+c_2\cos(x)+c_3\tan(x)=c_2$$and therefore $c_2=0$. On the other hand$$x=\frac\pi2\implies c_1\sin(x)+c_3\tan(x)=\frac{c_1}{\sqrt2}+c_3=0$$and therefore $c_3=-\frac{c_1}{\sqrt2}$. Finally, see what happens when $x=\frac\pi6$, in order to deduce that $c_1=c_3=0$.

$\endgroup$
4
  • 1
    $\begingroup$ Apart from this I wanna know that how is the statement that '' three vectors are linearly independant if they are non coplanar'' true. I think all of the functions are lying on the same plane. $\endgroup$
    – user187604
    Aug 24, 2018 at 9:25
  • $\begingroup$ Your question is about Linear Algebra. What is the vector space that you are working with? $\endgroup$ Aug 24, 2018 at 10:03
  • $\begingroup$ well I don't know the perfect tags. You are welcome to retag it. $\endgroup$
    – user187604
    Aug 24, 2018 at 10:07
  • 1
    $\begingroup$ I am not talking about tags. You mentioned linear independence. So, which vector space do you have in mind? $\endgroup$ Aug 24, 2018 at 10:18
2
$\begingroup$

Suppose that $c_1 \sin x + c_2 \cos x + c_3 \tan x =0$ for all $x \in ( - \pi/2, \pi/2)$.

With $x=0$ we get $c_2=0$. Hence $ \sin x (c_1+\frac{c_3}{\cos x})=0$ for all $x \in ( - \pi/2, \pi/2).$.

If $x \ne 0$ we derive $c_1+\frac{c_3}{\cos x}=0$ .

It is your turn to show that $c_1=c_3=0$.

$\endgroup$
1
  • 1
    $\begingroup$ Apart from this I wanna know that how is the statement that '' three vectors are linearly independant if they are non coplanar'' true. I think all of the functions are lying on the same plane. $\endgroup$
    – user187604
    Aug 24, 2018 at 9:25
2
$\begingroup$

Those 3 functions are linearly independent. Plugging $x=0$, you get that $c_2=0$. Now as $x \to \pi/2$, $\sin$ is bounded while $\tan$ is not, therefore $c_3=0$. Finally this imposes $c_1=0$ as $\sin$ is not always vanishing.

$\endgroup$
3
  • 1
    $\begingroup$ Apart from this I wanna know that how is the statement that '' three vectors are linearly independant if they are non coplanar'' true. I think all of the functions are lying on the same plane. $\endgroup$
    – user187604
    Aug 24, 2018 at 9:24
  • $\begingroup$ @user187604 because you mix vector with a function whose values are vectors. For any given x (except those where tan does not exists) the 3 vectors produced are linearly dependent. But the factors will change with x changing. $\endgroup$
    – Ister
    Aug 24, 2018 at 11:04
  • 1
    $\begingroup$ @user187604 "I think all of the functions are lying on the same plane." Well, they're not. $\endgroup$
    – Jack M
    Aug 24, 2018 at 12:29
1
$\begingroup$

Let $x$ tend to $\pi\over 2$ form below. Then $\tan x$ grows up to $\infty$ while $\sin x$ and $\cos x$ are both bounded. This leads to $c_3=0$. Also let $x=0$ and $x=\dfrac{\pi}{2}$ which yields to $c_2=0$ and $c_1=0$ respectively arguing that these three functions are linearly independent.

$\endgroup$
3
  • 1
    $\begingroup$ Apart from this I wanna know that how is the statement that '' three vectors are linearly independant if they are non coplanar'' true. I think all of the functions are lying on the same plane. $\endgroup$
    – user187604
    Aug 24, 2018 at 9:24
  • $\begingroup$ Well! This is a virtual definition for example you can consider $\sin x=(1,0,0)\\\cos x=(0,1,0)\\\tan x=(0,0,1)$ over the space of functions $\{f(x)|f(x)=c_1\sin x+c_2\cos x+c_3\tan x,c_1,c_2,c_3\in\Bbb R\}$ $\endgroup$ Aug 24, 2018 at 9:29
  • $\begingroup$ Even on the $\Bbb R^2$ the dimension can be $\infty$. Let the space of analytic functions. Each analytic function can be expressed as sum of $x^n$ s multiplied by some real factor called Taylor series coefficients. Since we have infinitely many such coeffs so would have infinite dimension (this is because Taylor series coeffs are unique). $\endgroup$ Aug 24, 2018 at 9:32
0
$\begingroup$

The confusion here is in the word "coplanar".

The functions are not elements of the $xy$-plane. $\sin$, $\cos$, and $\tan$ live in an infinite-dimensional vector space. Even though we draw them as graphs in a plane, they are not elements of the plane, since the plane is just a set of points.

$\endgroup$

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