<s>[INST] Power Series proofs $$\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}$$
$$\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}$$
$$\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}$$
Show that $\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)$ for every $x, y \in\mathbb R$.
Show that $\alpha(x)^3 + \beta(x)^3 + \gamma(x)^3 − 3\alpha(x)\beta(x)\gamma(x) = 1$ for every $x\in\mathbb R$.
My work:
Ive noticed that $$\sinh(x)=\sum_{j=0}^\infty \frac{x^{2j+1}}{(2j+1)!}$$
and $$\cosh(x)=\sum_{j=0}^\infty \frac{x^{2j}}{(2j)!}$$
Can hyperbolic trig identies be used to solve these and if so how?
 [/INST] For  the latter, notice that \begin{align*} \alpha' &= \beta, \\ \beta' &= \gamma, \\ \gamma' &= \alpha. \end{align*} Then we see \begin{align*}\frac d {dx} (\alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma ) &= 3\alpha^2 \alpha'+ 3\beta^2 \beta' + 3\gamma^2 \gamma' - 3\alpha' \beta \gamma - 3\alpha \beta' \gamma - 3\alpha \beta \gamma' \\
&= 3\alpha^2 \beta + 3\beta^2 \gamma + 3\gamma^2\alpha - 3\beta^2 \gamma - 3\gamma^2 \alpha - 3\alpha^2\beta = 0.
  \end{align*} Thus $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma$ is constant. Then $\alpha(0) = 1, \beta(0) = \gamma(0) = 0$ shows that $$\alpha(x)^3 + \beta(x)^3 + \gamma(x)^3 - 3\alpha(x) \beta(x) \gamma(x) = 1, \,\,\,\, \forall x \in \mathbb R.$$
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