I have tried finding a Taylor series to evaluate the sum $\sum_{n=0}^{\infty}\frac{1}{(3n)!}$, but am unable to find such a function. Is there a Taylor Series that could frame this sum?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Start with $e^x=\sum_n x^n/n!$ and Look at $g(x)=e^{x}+e^{wx}+e^{w^2x}$ for $w=e^{2\pi i/3}$. What happens with the terms with exponent not divisible by $3$? $\endgroup$– conditionalMethodCommented Nov 26, 2019 at 17:18
-
$\begingroup$ @PeterForeman This is only useful given the magical knowledge of the closed-form for this power series. $\endgroup$– Clement C.Commented Nov 26, 2019 at 17:33
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Hint: Find a differential equation for $$ f(x) = \sum_{n=0}^\infty \frac{x^{3n}}{(3n)!} $$ Differentiate this 3 times and relate it to the original.
-
$\begingroup$ This is the most general/systematic way to tackle such questions, provided some familiarity with solving differential equations. $\endgroup$ Commented Nov 26, 2019 at 17:29
-
1$\begingroup$ @ClementC. By what measure? That is just an empty claim. It is just one approach. That's all. $\endgroup$ Commented Nov 26, 2019 at 18:14
-
$\begingroup$ It's a very general approach which can be used for many things of the form $\sum_{n=0}^\infty a_n$, while many other approaches don't. That's all. $\endgroup$ Commented Nov 26, 2019 at 18:21
$\begingroup$
$\endgroup$
1
If one has a function given by a convergent power series $$f(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+\cdots$$ then the sum of every third term $$g(x)=\sum_{n=0}^\infty a_{3n}x^{3n}=a_0+a_3x^3+a_6x^6+a_9x^9+\cdots$$ may be obtained by the trick of "series trisection".
Write $\newcommand{\om}{\omega}\om=\exp(2\pi i/3)=\frac12(-1+i\sqrt3)$, so that $\om^3=1$, and $1+\om+\om^2=0$. Then $$f(\om x)=a_0+\om a_1x+\om^2 a_2x^2+a_3x^3+\cdots$$ and $$f(\om^2 x)=a_0+\om^2 a_1x+\om a_2x^2+a_3x^3+\cdots.$$ Adding these to each other, and to $f(x)$ gives $$f(x)+f(\omega x)+f(\omega^2x)=3a_0+3a_3x^3+3a_6x^6+\cdots=3g(x).$$
In your application, you want to compute $g(1)$ where $f(x)=\exp(x)$.
answered Nov 26, 2019 at 17:19
-
$\begingroup$ This trick is also known as a root-of-unity filter, on the grounds that $1^n+\omega^n+\omega^{2n}=0$ if $n$ isn't a multiple of 3. $\endgroup$ Commented Nov 26, 2019 at 17:30