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The problem is:

Show that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set of functions defined on $\mathbb{R}$.

The problem expects the student to use a computer program such as Matlab.

To solve the problem I created a matrix in Matlab with $t$ ranging from $1$ to $7$, then row reduced it and got the identity matrix as result. I.e. the columns of the matrix is linearly independent.

Is it enough to just show for $t=1\to 7$; could it not possibly break down at some other number? And is there a more elegant way of solving this without the use of "brute force"?

Here's the Matlab code if necessary:

for t = 1:7
A(t,:) = [1 cos(t) (cos(t))^2 (cos(t))^3 (cos(t))^4 (cos(t))^5 (cos(t))^6];
end
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It will break down at some other numbers (extreme example $t=\pi, 2\pi, \dots, 7\pi$, but there are more subtle ones). But if you can find some collection of seven $t$'s, that settles it. There is a much easier (for me) way that involves no Matlab. –  André Nicolas Oct 8 '11 at 20:29
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Simply composing each of the functions with $\cos^{-1}$ on the right gives 7 polynomials on [-1,1], which must be independent because they have different degree. Since function composition preserves linear combinations, the original functions must also be independent. –  Henning Makholm Oct 8 '11 at 20:48
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Those polynomials Henning is referring to are termed Chebyshev polynomials. –  J. M. Oct 9 '11 at 2:08
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4 Answers

up vote 17 down vote accepted

The computational approach that you used works well. For our particular collection of functions, or more generally for the powers of a single function $f(t)$, there is a straightforward non-computational approach. As a bonus, the argument below works just as well for $666$ as it does for $6$.

Suppose to the contrary that our collection $\{1,\cos t, \cos^2 t, \dots, \cos^6 t\}$ of functions is not linearly independent. Then there are constants $a_0, a_1, \dots, a_6$, not all $0$, such that $$a_0+a_1 \cos t +a_2 \cos^2 t +\cdots +a_6 \cos^6 t=0 \quad\text{ for all $t$}.$$

Let $P(x)=a_0+a_1x+a_2x^2+\cdots +a_6x^6$. Not all the coefficients $a_i$ are $0$, so the equation $P(x)=0$ has at most $6$ solutions. (A non-zero polynomial of degree $\le n$ has at most $n$ roots.) Thus if we can show that the function $\cos t$ can take on more than $6$ different values, we obtain the desired contradiction.

But $\cos t$ takes on infinitely many different values as $t$ ranges over the reals, or indeed over any interval of non-zero length. This completes the proof.

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Thank you for your answer, André! Forgive me for asking, but it is not obvious to me why the fact that $\cos t$ can take on more than 6 different values makes a contradiction? –  Jodles Oct 10 '11 at 23:47
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We are supposing that there are $a_i$ not all $0$ such that $P(\cos t)=0$, where $P(x)$ is the polynomial of my post. Since $\cos t$ can take on more than $6$ values, that would mean that $P(x)$ would be $0$ at more than $6$ places. But a non-zero polynomial of degree $\le 6$ can be $0$ at at most $6$ places. The only way out is to conclude that $a_i=0$ for all $i$, meaning that your $7$ functions are linearly independent. For the same reason, $1$, $\tan t$, $\tan^2 t$, and so on up to $\tan^{101} t$ is a linearly independent collection. –  André Nicolas Oct 11 '11 at 1:03
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Two comments:

  1. Suppose that you have functions $f_1(t)$, $f_2(t),\ldots,f_k(t)$, and that they are linearly dependent in the vector space of all real-valued functions. That means that you can find real numbers $\alpha_1,\ldots,\alpha_k$ such that $$\alpha_1 f_1(t)+\cdots + \alpha_k f_k(t) = 0.$$ That means that for any value $a$ of $t$, you will have $$\alpha_1f_1(a) + \cdots + \alpha_k f_k(a) = 0.$$ In particular, if you pick $k$ different values for $t$, $a_1,\ldots,a_k$, then you have: $$\begin{align*} \alpha_1f_1(a_1) + \cdots + \alpha_k f_k(a_1) &= 0\\ \alpha_1f_1(a_2) + \cdots + \alpha_k f_k(a_2) &= 0\\ &\cdots\\ \alpha_1f_1(a_k) + \cdots + \alpha_k f_k(a_k) &= 0 \end{align*}$$ which in turn means that: $$\alpha_1 \left(\begin{array}{c}f_1(a_1)\\f_1(a_2)\\ \vdots \\ f_1(a_k)\end{array}\right) +\alpha_2\left(\begin{array}{c}f_2(a_1)\\f_2(a_2)\\\vdots\\f_2(a_k)\end{array}\right) + \cdots + \alpha_k\left(\begin{array}{c}f_k(a_1)\\f_k(a_2)\\\vdots\\f_k(a_k)\end{array}\right) = \left(\begin{array}{c}0\\0\\\vdots\\0\end{array}\right),$$ so the columns of the corresponding matrix are linearly dependent, so the corresponding matrix is singular.

    By contrapositive, if the matrix you get by evaluating the functions at $k$ different points as you did is nonsingular, then the functions have to be linearly independent, as you conclude.

    However, the converse is not true: it may be that you get "unlucky" and evaluate at points where the matrix is singular, even if the functions are not. If instead of using $t=1,2,\ldots,7$ you had used $t=\pi,2\pi,\ldots,7\pi$, your matrix would have been singular, even though the functions are, as you concluded, linearly independent. So the matrix being nonsingular is sufficient, but it is not necessary for the functions to be linearly independent.

  2. As to other methods of doing this... (comment two): the standard way when your functions can be differentiated enough times (as they can be here), is to use the Wronskian.

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Of course, compute the Wronskian. –  GEdgar Oct 8 '11 at 21:14
    
Thank you for your answer, Arturo! I think my problem might stem from a misunderstanding of what linear independence requires. I thought for functions to be linearly independent, they need to be so for all $\mathbb{R}$; however, it seems like it's the other way around: i.e. if a set of functions are not linearly dependent for all $\mathbb{R}$, then they are linearly independent. So if we can show one example of linear independence, then we're done. Is that the gist of what you were saying? –  Jodles Oct 10 '11 at 23:51
    
@Jodles: I honestly don't understand your comment. By definition, there are only two possibilities for any set of vectors: either they are linearly independent, or they are linearly dependent. It is always the case that if a set of vectors is not linearly dependent, then it is linearly independent; and that if it is not linearly independent, then it is linearly dependent. For functions, equality means equality as functions, so when we write that $a_1f_1(t)+\cdots+a_nf_n(t)=0$, we mean that this holds for all values of $t$. –  Arturo Magidin Oct 11 '11 at 4:12
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The equality $$ \cos x=\frac{e^{ix}+e^{-ix}}{2} $$ shows that $(\cos x)^n$, being a polynomial of degree $n$ in $e^{ix}$ and $e^{-ix}$, is not a linear combination of lower powers of $\cos x$.

The fact (implicitly used) that the functions $e^{inx}$ are linearly independent follows from the equality $$ \frac{d}{dx}\ e^{inx}=in\ e^{inx}. $$

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The system of functions $f_i,i=1,2,...,n$ is linearly independent on some set $A$ iff the condition

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \displaystyle \sum\limits_{i}\alpha_i f_i(x) = 0$ for all $x\in A$

is satisfied only for $\alpha_i = 0$ for all $i=1,...,n$. So, you should solve as many equations, as there are elements in $A$. On the other hand, it's sufficient to take just some elements from $A$ to show that $$ \sum\limits_{i}\alpha_i f_i(x_j) = 0 $$ for all $j=1,...,m$ implies that $\alpha_i = 0$ for all $i$.

I would say that in your case it's sufficient to consider $8$ points $x_j$ to obtain the over-determined system of linear equations on $\alpha_i$ since $n=7$ in your case and to obtain the over-determined system usually it is sufficient to have $n+1$ equation.

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