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I found this interesting exercise on a calculus book (Stewart)

Let $$ u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\cdots $$ $$ v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\cdots $$ $$ w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots $$ Show that $$u^3+v^3+w^3-3uvw=1$$ It turns out to be an interesting application of the 3rd root of unity, which greatly simplified the (could be) tedious calculation. I wonder if it has any deeper interpretation. (At least I don't see how to easily generalize it.) Can anybody explain this? (I don't need help on solving this problem.)

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This is a good question. –  Bombyx mori Nov 29 '12 at 5:51
    
There are more relations.If $u(x)=\sum \limits_{k=0}^\infty \frac{x^{3k}}{(3k)!}$ then $u(2x)=3u^2(x)-2u(-x)$ and also $u(3x)=1+9u(x)v(x)w(x)$ .For more details, you can see my question math.stackexchange.com/questions/180832/… –  Mathlover Nov 29 '12 at 6:37
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2 Answers

up vote 7 down vote accepted

The background is the discrete Fourier transform. The LHS of the identity you wrote down is the determinant of a circulant matrix

$$\left[ \begin{array}{ccc} u & v & w \\\ w & u & v \\\ v & w & u \end{array} \right]$$

and the discrete Fourier transform tells you that its eigenvalues are

$$u + v + w, u + \omega v + \omega^2 w, u + \omega^2 v + \omega w$$

where $\omega$ is a primitive third root of unity, and hence that the product of these is the determinant. But

$$u + \omega^k v + \omega^{2k} w = \exp (\omega^k x)$$

so their product is

$$\exp (x + \omega x + \omega^2 x) = \exp(0) = 1.$$

The natural generalization involves larger circulant matrices and $\exp (\zeta^k x)$ for $\zeta$ a primitive $n^{th}$ root of unity. Note that when $n = 2$ you get the hyperbolic Pythagorean theorem.

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(This isn’t a generalisation, but this is perhaps a more elementary way of looking at the original problem.)

Looking at the three power series, each is the derivative of the next one, cycling around: $$ u = \frac{dv}{dx} \qquad v = \frac{dw}{dx} \qquad w = \frac{du}{dx} \qquad $$

So differentiating the expression given gives:

$\displaystyle \frac{d}{dx} (u^3 + v^3 + w^3 - 3uvw) $
$\displaystyle \qquad = 3u^2\frac{du}{dx} + 3v^2\frac{dv}{dx} + 3w^2\frac{dw}{dx} - 3(\frac{du}{dx}vw + \frac{dv}{dx}uw + \frac{dw}{dx}uv)$
$\displaystyle \qquad = 3u^2v + 3v^2w + 3uw^2 - 3(v^2w + uw^2 + u^2v)$
$\displaystyle \qquad = 0$

So $u^3 + v^3 + w^3 - 3uvw$ is constant with respect to $x$ (since its derivative is always $0$), and evaluating it at any particular value of $x$ gives its value for all $x$. Now choosing $x=0$, we get $u = 1$, $v=w=0$, and so $u^3 + v^3 + w^3 - 3uvw = 1$.

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I like this solution, it is quick! (although it shed no light on the background.) –  hyh Nov 29 '12 at 6:14
    
This may "shed no light on the background", but it does show how calculus can be used to achieve the answer. –  Mark Bennet Nov 29 '12 at 7:41
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It sheds plenty of light on the background. It clues you in to the fact that $u, v, w$ are all solutions of the differential equation $\frac{d^3}{dx^3} f = f$. –  Qiaochu Yuan Dec 2 '12 at 0:52
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