A series question about Taylor series

Question

$a=1+\frac{x^{3}}{3 !}+\frac{x^{6}}{6 !}+\cdots$

$b=\frac{x}{1 !}+\frac{x^{4}}{4 !}+\frac{x^{7}}{7 !} \cdots$

$c=\frac{x^{2}}{2 !}+\frac{x^{5}}{5 !}+\frac{x^{8}}{8 !}+\cdots$

Let:$ u=a^{2}+b^{2}+c^{2}, $

What is the value of $u(\ln 2)\ $?

Answer 1

As, for a cubic root of unity, $1+\omega+\omega^2=0$, you can sum $e^x,e^{\omega x},e^{\omega^2x}$ to obtain every third term of the exponential, times $3$; also $e^x,\omega e^{\omega x},\omega^2e^{\omega^2x}$ and $e^x,\omega^2 e^{\omega x},\omega e^{\omega^2x}$.

If you evaluate at $x=\log 2$, simplifying every sum $\omega^{0k}+\omega^{1k}+\omega^{2k}$, you get

$$(2+2^\omega+2^{\omega^2})^2+(2+\omega2^\omega+\omega^22^{\omega^2})^2+(2+\omega^22^\omega+\omega2^{\omega^2})^2=12+2\cdot3\cdot2^{\omega+\omega^2}=15$$

to be divided by $3^2$, i.e.

$$\frac53.$$

Answer 2

Note that: $a(x)=\sum_{k=0}\limits^{\infty}\frac{x^{3k}}{(3k)!},\ b(x)=a^{'}(x),c(x)=a^{''}(x),a(x)=a^{'''}(x),a(x)+b(x)+c(x)=e^x$

$a(x)+a^{'}(x)+a^{''}(x)=e^x$

a particular solution is $ a(x)=\frac{e^x}{3} $

$r^2+r+1=0,r_1=-\frac{1}{2}-\frac{\sqrt 3}{2}i,r_1=-\frac{1}{2}+\frac{\sqrt 3}{2}i,$

$s.t.$the general solution is$ a(x)=\frac{e^x}{3}+e^{-\frac{x}{2}}(Asin\frac{\sqrt 3}{2}x+Bcos\frac{\sqrt 3}{2}x)$

$a(0)=0,then\ B=-\frac{1}{3}$.next prime the a$(x)$ and $a^{'}(0)=0$,we can get $A$.

Now we get $a(x)$,easily to get $b(x)$ and $c(x)$,

so $ a\ (ln\ 2),b\ (ln\ 2)$ and $c\ (ln\ 2)$ all are known,$u(ln\ 2)$ is also known.

the result should be $ \frac{5}{3}. $