How to prove that
$$
\sum\limits_{n=0}^\infty \frac {x^{5n}} {(5n)!}=
\frac{2}{5} e^{-\cos \left( 1/5\,\pi \right) x}\cos \left( \sin
\left( 1/5\,\pi \right) x \right) +\frac{2}{5}\, e^{\cos \left( 2/5\,
\pi \right) x}\cos \left( \sin \left( 2/5\,\pi \right) x \right) +\frac{1}{5} e^{x}?
$$
Answer
Hint. Clearly
$$
\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}
$$
where $\omega=\mathrm{e}^{2\pi i/5}$, since
$$
\sum_{j=1}^5 \omega^{jn}=\left\{\begin{array}{ccc} 5&\text{if}& 5\mid n, \\ 0&\text{if} &5\not\mid n. \end{array}\right.
$$
But
$$
\omega=\cos (2\pi/5)+i\sin (2\pi/5), \,\,\omega^2=\cos (4\pi/5)+i\sin (4\pi/5),
\,\,\omega^3=\overline{\omega^2},\,\,\omega^4=\overline{\omega},
$$
and so
$$
\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}=\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\frac{1}{5}\left(\mathrm{e}^x+2\mathrm{e}^{x\cos(2\pi/5)}\cos\big(\sin(2\pi/5)\big)+2\mathrm{e}^{x\cos(4\pi/5)}\cos\big(\sin(4\pi/5)\big)\right).
$$
No comments:
Post a Comment