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I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.

The question reads:

Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove that the functions $x^{3}$, $\sin(x)$, and $\cos(x)$ are linearly independent.

My proof goes as follows:

For these to be linearly dependent there must exist an $a_{i} \neq0$, where $ i = 1, 2, 3$ such that $$a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x) = 0. $$ So, we'll do this in 3 cases:

Case 1: $x = 0$

In this case, $x^{3} = 0$, $\sin(x) = 0$ but $\cos(x) = 1$. So, we have $$0\times a_{1} + 0\times a_{2} + 1\times a_{3} = 0.$$ So, $a_{1}$ and $a_{2}$ could be anything but $a_{3}$ must be 0.

Case 2: $x \in (0,1)$

In this case, $x^{3} \neq 0$, $\sin(x) \neq 0$ and $\cos(x) \neq 0$. So, for this to be true, $a_{1}$, $a_{2}$ and $a_{3}$ all must be $0$.

Case 3: $x = 1$

In this case, $x^{3} = 1$, $\sin(x) = .8...$ and $\cos(x) = .5...$. So, we have $$1\times a_{1} +.8\times a_{2} + .5\times a_{3} = 0.$$

So, $a_{3}$ could be any value, while $a_{1}$ and $a_{2}$ must be $0$.

So, if $a_{1} \neq 0$ then we have a problem in Case 3. If $a_{2} \neq 0$ we have a problem in Case 3. If $a_{3} \neq 0$ we have a problem in Case 1. So, we know that all of the $a$ values must be $0$ and we complete the proof.

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    $\begingroup$ You're not really considering cases in the sense typically used in proofs, which means considering different logical possibilities to make sure you cover them all. Plugging in values just gives you more logical consequences of the equation rather than cases. $\endgroup$ Feb 15, 2017 at 3:59
  • $\begingroup$ How is this not exhaustive? Couldn't my three cases be rewritten in the logical form $(x = 0 \lor x \in (0,1) \lor x = 1)$ and then taken case by case? $\endgroup$ Feb 15, 2017 at 4:01
  • $\begingroup$ Thanks, I misunderstood your intention. You can evaluate your function wherever you want, and there is no need to consider every possibility, just to get enough info to conclude that the $a_i$s are zero. I've elaborated in an answer. $\endgroup$ Feb 15, 2017 at 4:11
  • $\begingroup$ If $f_0(x) = x^3, f_1(x) = sin(x), f_2(x) = cos(x)$ what you need to do is to find three values $x_0,x_1,x_2 \in [0,1]$ such that the vectors $\vec{f}_j = (f_0(x_j),f_1(x_j),f_2(x_j))$ for $j = 0,1,2$ such that they're linearly indipendent. The choice is not hard. $\endgroup$ Feb 15, 2017 at 10:02

5 Answers 5

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If you make no assumptions about $a_i$s and then show they must be zero, you have proved linear independence.

From the evaluation at $x=0$ (case 1) you correctly conclude that $a_3=0$. At this point, I suggest simplifying your expression to $a_1x^3+a_2\sin(x)$.

In your evaluation at $x\in(0,1)$ (case 2) you have a logical error of assuming that a sum of numbers being zero requires each number to be zero. $a+b=0$ does not imply $a$ and $b=0$

There is also an error here in the structure of your proof, because if you had shown that all of the $a_i$s are zero, you would be done already.

You have a similar logical error for the evaluation at $x=1$ (case 3).


It seems you are thinking of the outcome of the values of $a_i$s as depending on what $x$ is, hence the "cases" that are treated independently. But the $a_i$s are always the same, for all $x$. Your idea of evaluating at particular points is a good one, and they build on each other. The evaluation at $x=0$ shows that $a_3=0$. Evaluation at two other particular values of $x$, say, $x=\frac12$ and $x=1$, would give you two equations in two unknowns for $a_1$ and $a_2$, and you could show the only solution is $a_1=a_2=0$.

Alternatively, you could take the derivative of your expression to get $3a_1x^2+a_2\cos(x)=0$, and then evaluation at $x=0$ will give you $a_2=0$ directly, after which you'll be left with $a_1x^3\equiv 0$ from which it isn't hard to show $a_1=0$ either by evaluation at $x=1$ or taking the derivative $3$ times.

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Differentiation can be powerful in cases like this. BobaFret's suggestion of using the Wronskian is one possibility, but you can do it more directly. Your errors have been discussed well in Jonas Meyer's answer, and he also discussed differentiation.

I want to add why differentiation is a good idea. As you noticed, the condition is simplest to work with at $x=0$. You can of course take three points in $[0,1]$ and evaluate the condition at each one, but it is more convenient to work at the origin if at all possible. Since looking at only $x=0$ gives you just one condition, you should next look near the origin. This process of "looking at a condition near a point" is often means differentiating the condition there. It is often easier to differentiate a given condition at an easy point than to study the condition elsewhere. Even if there are several easy points you could pick, it can be beneficial to choose only one but look at different derivatives at the same point. This is not just a little calculus trick; this idea is not at all unusual in my research.


Here's a full solution with this approach:

You have a function $$ f(x)=a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x) $$ and you know that $f(x)=0$ for all $x\in[0,1]$. Evaluating at $x=0$ gives you $f(0)=a_3=0$, so in fact $$ f(x)=a_{1}x^{3} + a_{2}\sin(x). $$ Since $f(x)=0$ for all $x\in[0,1]$ and all functions are differentiable (on the whole real line), you also have $$ f'(x)=3a_1x^2+a_2\cos(x)=0 $$ for all $x\in[0,1]$. Evaluating this at zero leaves you $0=f'(0)=a_2$. Therefore $$ f(x)=a_{1}x^{3}. $$ Now you can study $f(1)$ to find that $a_1=0$. If you want to keep on differentiating, you can also inspect $f'''(0)$ and draw the same conclusion.

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Your case 2 and case 3 are wrong.

For case 2, just because $a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$ does not imply that the $a_i$ are all zero. Actually, for any $x \ne 0$, and any $a_2, a_3$, $a_1$ can always be chosen so that $a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$ by setting $a_{1}=-\dfrac{a_{2}sin(x) + a_{3}cos(x)}{x^{3}}$.

For your case 3, $\sin(1) \ne 1$ and $\cos(1) \ne 0$. If you use $\pi/2$, then this works.

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    $\begingroup$ But $\pi/2\not\in[0,1]$. $\endgroup$
    – Wojowu
    Feb 15, 2017 at 13:24
  • $\begingroup$ @BenVoigt I know that. The issue is that you can't plug in $x=\pi/2$. $\endgroup$
    – Wojowu
    Feb 15, 2017 at 18:44
  • $\begingroup$ Oh right, in case 3 he is trying to use that for $x$. $\endgroup$
    – Ben Voigt
    Feb 15, 2017 at 18:45
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Do you know what the Wronskian is? If so, this proof boils down to calculating a $3 \times 3$ determinant.

For $y_1 = x^3$, $y_2 = \sin x$, and $y_3 = \cos x$ the Wronskian is \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \\ \end{vmatrix}

which evaluates to a non-zero expression. This implies $y_1 = x^3$, $y_2 = \sin x$, and $y_3 = \cos x$ are linearly independent.

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Why not assume that $f(x) = a_1x^3+a_2 \sin x + a_3 \cos x$ is the zero function. Then note that $0 = f(0) = a_3$, so we may rewrite $f(x) = a_1x^3 + a_2 \sin x$. Evaluate this rewritten expression for $f$ at the point $\pi$ to show that $0 = f(\pi) = a_1 \pi^3 \Longrightarrow a_1 = 0$. Thus we can again rewrite $f(x) = a_2 \sin x$. Finally, evaluate this latest rewritten expression for $f$ at the point $\pi/2$ to find $0 = f(\pi/2) = a_2$.

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    $\begingroup$ I like this answer but, the vector space is V, functions on the interval [0,1] and $\pi,\pi/2>1$. $\endgroup$
    – cabo
    Feb 15, 2017 at 4:17
  • $\begingroup$ Your right! I overlooked that. $\endgroup$
    – joeb
    Feb 15, 2017 at 4:28

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