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Let $f(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}$ and $g(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n+1}}{(3n+1)!}$ and $h(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n+2}}{(3n+2)!}$

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

Today a calculus student asked me this question.
first thing that came in my mind that it is not true since if you take x=0 you will get 0=1, but someone pointed out to me that $f(0) $ will have $0^0$. Finally, finally I think I managed to solve it but under the assumptoin that $f(0)=1$, can you help to solve this question without any further assumptions.

Hint: may be it is usefull to notice that $h'=g$ and $g'=f$ and $f'=h$.

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marked as duplicate by Gerry Myerson, Potato, 5PM, Henry T. Horton, Micah Feb 12 '13 at 0:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Is that really $3n+1$, or should it be $3(n+1)$? (And same for the $+2$ term) – apnorton Feb 11 '13 at 22:35
You do not have to assume it. It is true that $f(0)=1$. The notation $x^0$ stands for $1$ here for all $x$. So no $0^0$ is involved. – 1015 Feb 11 '13 at 22:36
@anorton the question is stated correctly. it is (3n+1)! – i.a.m Feb 11 '13 at 22:41
@julien $x^0=1$ is true for all $x\ne 0$ but when substitution 0 in $f$ the first term will be $0^0$ – i.a.m Feb 11 '13 at 22:42
It looks "like" a hyperbolic identity $\cosh^2{x}-\sinh^2{x}=1,$ if we expand hyperbolic functions into its Taylor series $$\sinh{x}=\sum\limits_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}, \\ \cosh{x} = \sum\limits_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}. $$ – M. Strochyk Feb 11 '13 at 22:47

3 Answers 3

up vote 3 down vote accepted

The problem probably relies on the less known identity

$$a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-ac-bc) $$

Thus, we need to show


Now, let $u(x)=f^2+g^2+h^2-fg-fh-gh$. Then

$$u'=2ff'+2gg'+2hh'-f'g-fg'-fh'-f'h-gh'-g'h$$ $$=2fh+2gf+2hg-hg-ff-fg-hh-gg-fh=-u$$

this proves that


for some constant $C$.

Combining this we get: there exists a constant $C$ so that


Now setting $x=0$ we get $C=f(0)^3$ which seems to complete the proof.

I think that one can argue that this solution works over $\mathbb C$, and then setting $x=\omega$ a primitive third root of unity leads directly to $C=1$, which emphasizes that $f(0)=1$ is the only natural choice...


$f(0)=1$ is not an assumption. the definition of the Taylor series is

$$f(x)=\displaystyle \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}=1+\frac{x^{3}}{(3)!}+\frac{x^{6}}{(6)!}+...$$

Note that this is the same as

$$e^x= \sum_{n=0}^{\infty}\frac{x^{n}}{(n)!}$$

What is $e^0$?

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Thank you, it a nice solution :) – i.a.m Feb 11 '13 at 23:09

Easliy you can show that

$$g'(x)=f(x) \tag 1$$ $$h'(x)=g(x) \tag 2$$ $$f'(x)=h(x) \tag 3$$







You can see that



where c is a constant.


We know that for $x=0$


$c=-1$ Thus


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Differentiate the given expression with respect to $x$ and use the cyclic nature of the derivatives noted in the question. We obtain


Since $f'=h$, $g'=f$, and $h'=g$, this is zero. So the expression is a constant, and setting $x=1$ shows this constant is $1$.

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