series exercise asked by Imad zak in the Imad Zak math group

Exercise:

Let $h\left( x \right)=1+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{6}}}{6!}+....\,\,\,\,\,,\,\,\,\,\,q\left( x \right)=x+\frac{{{x}^{4}}}{4!}+\frac{{{x}^{7}}}{7!}+...\,\,\,\,,\,\,\,\,p\left( x \right)=\frac{{{x}^{2}}}{2!}+\frac{{{x}^{5}}}{5!}+\frac{{{x}^{8}}}{8!}+...$

Show that , ${{h}^{3}}\left( x \right)+{{q}^{3}}\left( x \right)+{{p}^{3}}\left( x \right)-3h\left( x \right)q\left( x \right)p\left( x \right)=1$

Solution: observe that $h\left( x \right)+q\left( x \right)+p\left( x \right)=1+x+\frac{{{x}^{2}}}{2!}+...={{e}^{x}}$

We have ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( {{a}^{3}}+{{b}^{3}} \right)+{{c}^{3}}$

And we know that ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$

So ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)+{{c}^{3}}={{\left( a+b \right)}^{3}}+{{c}^{3}}-3ab\left( a+b \right)$

But ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)$

So ${{\left( a+b \right)}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{\left( a+b \right)}^{2}}+{{c}^{2}}-c\left( a+b \right) \right)$

Hence ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-bc-ac \right)-3ab\left( a+b \right)$

So ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-ac-bc \right)-3ab\left( a+b \right)-3abc$

                           $=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-bc-ac \right)-3ab\left( a+b+c \right)$

                          $=\left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-bc-ac-3ab \right]$

                          $=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$

Thus ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ac-bc \right)$

So ${{h}^{3}}\left( x \right)+{{q}^{3}}\left( x \right)+{{p}^{3}}\left( x \right)-3h\left( x \right)q\left( x \right)p\left( x \right)$

$=\left( h\left( x \right)+q\left( x \right)+p\left( x \right) \right)\left( {{h}^{2}}\left( x \right)+{{q}^{2}}\left( x \right)+{{p}^{2}}\left( x \right)-h\left( x \right)q\left( x \right)-h\left( x \right)p\left( x \right)-q\left( x \right)p\left( x \right) \right)$

$={{e}^{x}}\left( {{h}^{2}}\left( x \right)+{{q}^{2}}\left( x \right)+{{p}^{2}}\left( x \right)-h\left( x \right)q\left( x \right)-h\left( x \right)p\left( x \right)-q\left( x \right)p\left( x \right) \right)$

Let $z\left( x \right)={{h}^{2}}\left( x \right)+{{q}^{2}}\left( x \right)+{{p}^{2}}\left( x \right)-h\left( x \right)q\left( x \right)-h\left( x \right)p\left( x \right)-q\left( x \right)p\left( x \right)$

So $z'\left( x \right)=2\left( hh'+qq'+pp' \right)-\left( h'q+q'h \right)-\left( h'p+p'h \right)-\left( q'p+p'q \right)$

But $\frac{d}{dx}\left( h\left( x \right) \right)=0+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{5}}}{5!}+....=p\left( x \right)$ , $\frac{d}{dx}\left( q\left( x \right) \right)=1+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{6}}}{6!}+....=h\left( x \right)$

And $\frac{d}{dx}\left( p\left( x \right) \right)=x+\frac{{{x}^{4}}}{4!}+....=q\left( x \right)$

So $z'\left( x \right)=2\left( hp+qh+pq \right)-\left( pq+{{h}^{2}} \right)-\left( {{p}^{2}}+qh \right)-\left( hp+{{q}^{2}} \right)$

           $=\left( hp+qh+pq \right)-\left( {{h}^{2}}+{{p}^{2}}+{{q}^{2}} \right)=-z\left( x \right)$

Since $$z'\left( x \right)=-z\left( x \right)\Leftrightarrow \frac{dz}{dx}=-z\Leftrightarrow \int{\frac{dz}{z}}=-\int{dx}\Leftrightarrow \ln \left| z \right|=-x+c\Leftrightarrow z\left( x \right)=c{{e}^{-x}}$$

Observe that $h\left( 0 \right)=1\,\,,\,\,q\left( 0 \right)=0\,\,\And \,\,p\left( 0 \right)=0\Leftrightarrow z\left( 0 \right)=1$ hence $c=1$

Therefore ${{h}^{3}}\left( x \right)+{{q}^{3}}\left( x \right)+{{p}^{3}}\left( x \right)-3h\left( x \right)q\left( x \right)p\left( x \right)={{e}^{x}}{{e}^{-x}}={{e}^{0}}=1$              Q.E.D

Or: by differentiating both sides by $x$ to get

So $3{{h}^{2}}h'+3{{q}^{2}}q'+3{{p}^{2}}p'-3\left( h'\left( qp \right)+\left( qp \right)'h \right)=0$

$\Leftrightarrow 3{{h}^{2}}q+3{{q}^{2}}h+3{{p}^{2}}q-3\left( {{p}^{2}}q+\left( q'p+p'q \right)h \right)=3{{h}^{2}}q+3{{q}^{2}}h+3{{p}^{2}}q-3\left( {{p}^{2}}q+{{h}^{2}}p+{{q}^{2}}h \right)=0$

so $f'\left( x \right)=0\Leftrightarrow f\left( x \right)=c$

Now to find the value of $c$ plug $x=0$ to get $c=1$                                   Q.E.D



*____________________________
first Method credit to شحات جامع‎
2nd Method credit to Kunihiko Chikaya