How can I find the general formula for the following real sequence

Question

How can I find the general formula for the following real sequence $$(x_n)_{n \ge0}=(1,0,-1,0,\frac{1}{2},0,\frac{-1}{6},0,\frac{1}{24},0,\frac{-1}{120},\ldots)$$ I just know $x_0$ to $x_{10}$ so how can I find the general formula for this real sequence?

Answer 1

It appears that you have $$x_n=\begin{cases} 0 & n\text{ odd}\\ (-1)^{n/2}\frac{1}{(n/2)!} & n\text{ even}\end{cases}$$

Answer 2

HINT: You will need to split the formula into two parts, one for odd subscripts and one for even subscripts. The one for odd subscripts should be pretty obvious. For the even subscripts, note that $x_{2n}$ is negative when $n$ is odd and positive when $n$ is even; you should know a simple function of $n$ that is $-1$ when $n$ is odd and $1$ when $n$ is even, and you can make this function a factor in your formula. Finally, you should recognize the denominators $1,1,2,6,24,120,\ldots$ as a familiar sequence; to give you a little more help, the next two terms are $720$ and $5040$.

Answer 3

x(n)=((-1)^(n/2))*(1/((n/2)!)) , if n=even x(n)=0 , if n=odd

Answer 4

To get it into one equation, you can use that $$ \cos{\tfrac{1}{2}n\pi} = \begin{cases} 1 & n = 4k \\ 0 & n= 4k \pm 1 \\ -1 & n=4k+2 \end{cases}, $$ where $k \in \mathbb{Z}$. This then gives the squence as $$ x_n = \frac{\cos{\frac{1}{2}n\pi}}{(n/2)!} $$ (What is $(n/2)!$ when $n$ is odd? You don't need to know for this, but you can get it from $(-1/2)!=\sqrt{\pi}$, which you'll learn about later on. Point is, it's not zero, so the odd terms are still zero.

Answer 5

Inspired by the solution by Chappers above, here's my contribution:

$$\large x_n=\frac{\Re (i^n)}{\big(\frac n2\big)!}$$