# Showing that $f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1$ for functions $f$, $g$, and $h$ defined by certain power series [duplicate]

I'm having trouble with this question, I have found the interval of convergence of $h(x)$ to be $(-\infty, \infty)$, but I don't know how to use that for the question as well as the hint. Any help would be appreciated. Thanks!

If $$f(x) =\sum_{n = 0}^\infty \frac{x^{3n}}{(3n)!},\qquad g(x) =\sum_{n = 0}^\infty \frac{x^{3n+1}}{(3n+1)!},\qquad h(x) = \sum_{n = 0}^\infty \frac{x^{3n+2}}{(3n+2)!}$$ show that $$f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1.$$ Hint: show that $h'(x) = g(x)$.

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## marked as duplicate by Qiaochu YuanFeb 12 '13 at 4:11

Are you saying you don't know how to prove the hint? Once you have the hint, you can prove something similar for $g'$ and for $f'$, and then you can differentiate the left side of the equation you're asked to prove, and see what happens. – Gerry Myerson Feb 10 '13 at 1:48
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Feb 10 '13 at 1:49
No I can prove the hint, I just need help with the "show that" part and thanks for the edit! – user61653 Feb 10 '13 at 1:54
Thank you! I got it now! Just didn't know what to do with the derivatives – user61653 Feb 10 '13 at 2:05

Note that $f'(x) = h(x)$, $g'(x)=f(x)$, and $h'(x)=g(x)$. Then consider

$$\frac{d}{dx} [f^3 + g^3 + h^3 - 3 f g h]$$

and show that it is zero using the above derivatives. Viz.,

\begin{align}\frac{d}{dx} [f^3 + g^3 + h^3 - 3 f g h] &= 3 f^2 f'+3 g^2 g'+3 h^2 h' - 3 f'gh-3 f g'h - 3 f g h'\\ &= 3 f^2 h+3 g^2 f + 3 h^2 g - 3 f'gh-3 f g'h - 3 f g h'\\ &= 0 \\\end{align}

Also note that $f(0)=1$, $g(0)=0$, and $h(0)=0$. Thus the integration constant of the above equation is 1, and

$$f^3(x) + g^3(x) + h^3(x) - 3 f(x) g(x) h(x)=1$$

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Aren't we trying to show that it is equal to 1 though? – user61653 Feb 10 '13 at 1:56
Yep. See the rest of the post (sorry it was fragmented). – Ron Gordon Feb 10 '13 at 2:00
No worries, and thank you! It makes sense! I just didn't know what to do when I had the derivatives! – user61653 Feb 10 '13 at 2:04

Notice that $h'=g$ and $g'=f$ and $f'=h$, now consider $$(fgh)'=f(g'h+gh')+f'gh$$ $$=fg'h+fgh'+f'gh$$ $$=f^2h+fg^2+gh^2$$ whcih implies that $$fgh=\int f^2h+fg^2+gh^2$$$$=\frac{f^3}{3}+\frac{g^3}{3}+\frac{h^3}{3}+c$$ which implies that $$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=c$$ for all $x\in \mathbb R$, substitute x=0 to get that c=1 and hence you get the result.

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