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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i.a.m
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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
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2You 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. – Julien Feb 11 '13 at 22:36
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@anorton the question is stated correctly. it is (3n+1)! – i.a.m Feb 11 '13 at 22:41
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@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
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No, in a Taylor series, the term $x^0$ (if it appears) means the function constant equal to $1$. Just like with polynomials. – Julien Feb 11 '13 at 22:44
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@julien Ok, then I don't need to assume that f(0)=1 it has to be 1 – i.a.m Feb 11 '13 at 22:46
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3It 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
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@M.Strochyk but the powers are of 3n and 3n+1 not 2n and 2n+1 – i.a.m Feb 11 '13 at 22:48
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I mean some similarity between $\sinh, \cosh$ with its series expansion and functions described in the question – M. Strochyk Feb 11 '13 at 23:04
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We just had this question, a day or two ago. – Gerry Myerson Feb 11 '13 at 23:09
3 Answers
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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
$$f^2+g^2+h^2-fg-fh-gh=\frac{1}{f+g+h}=e^{-x}$$
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
$$u(x)=Ce^{-x}$$
for some constant $C$.
Combining this we get: there exists a constant $C$ so that
$$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=C$$
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...
P.S.
$f(0)=1$ 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 $e^0$?
N. S.
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Easliy you can show that
$$g'(x)=f(x) \tag 1$$ $$h'(x)=g(x) \tag 2$$ $$f'(x)=h(x) \tag 3$$
$U(x)=f(x)g(x)h(x)$
$U'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$
$U'(x)=h^2(x)g(x)+f^2(x)h(x)+g^2(x)f(x)$
$P(x)=f^3(x)+g^3(x)+h^3(x)$
$P'(x)=3(f^2(x)f'(x)+g^2(x)g'(x)+h^2(x)h'(x))$
$P'(x)=3(f^2(x)h(x)+g^2(x)f(x)+h^2(x)g(x))$
You can see that
$3U'(x)=P'(x)=3(h^2(x)g(x)+f^2(x)h(x)+g^2(x)f(x))$
$3U(x)=P(x)+c$
where c is a constant.
$3f(x)g(x)h(x)=f^3(x)+g^3(x)+h^3(x)+c$
We know that for $x=0$
$f(0)=1$,$g(0)=0$,$h(0)=0$
$c=-1$ Thus
$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=1$
Mathlover
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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
$$3f^2f'+3g^2g'+3h^2h'-3(ghf'+fg'h+fgh').$$
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 $1$.
Potato
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