5

Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$

let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that

$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$

are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)

I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $

$$ \begin{vmatrix} f & g & h \\ f' & g' & h' \\ f'' & g'' & h'' \\ \end{vmatrix} $$ If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and $h(x)$ are linearly independent.

If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$

My attempt

Let

$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$

$W(f, g, h) =$ $$ \begin{vmatrix} \sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\ \cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\ -\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\ \end{vmatrix} $$

$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)] - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$

$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$

By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$

Not sure if this argument is sound?

Michael Hardy
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Zhoe
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    The very last equality is incorrect because you confused sums and differences. – anon Nov 11 '13 at 04:55
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    Without calculus, the determinent above is $0$ since the third ligne $L_3=-L_1$ ($L_1,L_3$ first and third line of the matrix. – Mohamed Nov 11 '13 at 05:17

4 Answers4

11

A short method :

Since $$\sin(x+\alpha)=(\cos \alpha) \sin x + (\sin \alpha) \cos x$$ we have $\sin(x+\alpha) \in \text{Span}\{\sin x, \cos x\}$

This implies that the rank of the family $(\sin(x+\alpha),\sin(x+\beta),\sin(x+\gamma))$ is $2$. That means that there are linearly dependants.

Mohamed
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3

As Mohamed pointed out in his answer, we can use the angle addition formula for $\sin$, viz.:

$\sin (x + \alpha) = \cos \alpha \sin x + \sin \alpha \cos x, \tag{1}$

$\sin (x + \beta) = \cos \beta \sin x + \sin \beta \cos x, \tag{2}$

$\sin (x + \gamma) = \cos \gamma \sin x + \sin \gamma \cos x, \tag{3}$

to see that all three functions lie in the subspace generated by $\sin x$ and $\cos x$, which is of course $\text{Span} \{ \sin x, \cos x \}$, and since $\sin (x + \alpha)$, $\sin (x + \beta)$, $\sin (x + \gamma)$ are thus three vectors in a subspace of dimension two, a linear dependence between them must exist.

Going a step further, we can actually solve for $\sin x$, $\cos x$ in terms of $\sin (x + \alpha)$, $\sin(x + \beta)$ and thus find the linear dependence explicitly. We have, from (1) and (2),

$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = \begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}; \tag{4}$

the inverse of

$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \tag{5}$

is readily seen to be

$(\cos \alpha \sin \beta - \sin \alpha \cos \beta)^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}, \tag{6}$

from which it follows that

$\begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}\begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}. \tag{7}$

(7) gives $\sin x$, $\cos x$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; the resulting formulas may then be inserted into (3) to express $\sin(x + \gamma)$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; I leave the necessary algebra to my readers. One must of course remain vigilant to the possibility that $\sin(\beta - \alpha) = 0$; but this is a special case best handled on its own.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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2

Explain how you got that expression for $W(f,g,h)$ equal to $0$. Otherwise the argument is good.

Another idea: consider rewriting the functions with the sum-angle formulas, then interpret the three functions as three vectors in a two-dimensional vector space spanned by $\sin$ and $\cos$.

anon
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  • This expression $= -sin(x + \alpha)[sin((x + \gamma) +(x + \beta))] + sin(x + \beta)[sin((x + \gamma) + (x + \alpha))] - sin(x + \gamma)[sin((x + \beta) + (x + \alpha))] = 0$? @anon – Zhoe Nov 11 '13 at 05:01
  • @zhoe Yes that one. (Except you need to fix it because it's incorrect. See my comment above.) – anon Nov 11 '13 at 05:06
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Other solution:

If you know that the set of diffrential equation: $$(E) \quad y''+y=0$$ solutions is a two dimensinal space, then since the three applications : $x \mapsto \sin(x+a)$ where $a \in\{\alpha,\beta,\gamma \}$ are solutions of $(E)$, there are linearly dependantes.

Mohamed
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