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I recently saw the question What is the closed form for $S=\displaystyle\sum_{n=1}^{\infty} \dfrac{\sin ({n})}{n!}$?

I became interested in the problem and generalized the equation using the same method provided in the link.

$$\displaystyle\sum_{n=0}^{\infty} \dfrac{\sin ({kn})}{n!} = e^{\cos(k)}\sin(\sin(k)),$$ $$\displaystyle\sum_{n=0}^{\infty} \dfrac{\cos ({kn})}{n!} = e^{\cos(k)}\cos(\sin(k)),$$ $$\displaystyle\sum_{n=0}^{\infty} \dfrac{\sin^2 ({kn})}{n!} = \frac{e+e^{\cos(2k)}\cos(\sin(2k))}{2},k\in\mathbb{R}.$$ Are there any other methods besides complex analysis that can provide a closed form for $\displaystyle\sum_{n=0}^{\infty} \dfrac{\sin ({n})}{n!}$?

Villa
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Larry
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1 Answers1

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I tried alternative methods to find the bijection between complex numbers and some operation or transformation. First thing I thought about was that if we call the derivative operator $D$, we have that:

$$D^2\cos{t}=-\cos{t} $$ $$D^2\sin{t}=-\sin{t} $$

Therefore in this context we have that $D^2=-I$, so I started looking for differential equations. I called the functions:

$$y(t)=\sum_{n=0}^\infty \frac{sin{(tn)}}{n!}$$ $$x(t)=\sum_{n=0}^\infty \frac{cos{(tn)}}{n!}$$

Then I derived the functions and fount a relation. Now the problem is to solve the following system with initial conditions $x(0)=e \text{, } \space y(0)=0$ and evaluate $y(t)$ in $t=1$:

$$y'=x\cos{t}-y\sin{t}$$ $$x'=-y\cos{t}-x\sin{t}$$

Using the subs $x=R(t)\cos(\theta (t))$ and $y=R(t)\sin(\theta (t))$ we get the answer:

$$y(t)=e^{\cos{t}}\sin(\sin{t})$$ $$x(t)=e^{\cos{t}}\cos(\sin{t})$$

Villa
  • 914