# What is the general term $(a_n)$ of the alternating sequence $\cos(3n \pi/2)$?

What is the general term $(a_n)$ of the alternating sequence $\displaystyle \cos \left( \frac{3n \pi}{2} \right)$ from $1$ to $\infty$, $n \in \mathbb{N}$ ?

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1
Does t = n? Have you tried evaluating a few terms yourself? – Qiaochu Yuan Jan 4 '11 at 12:08
This is alternating, and furthermore it is periodic with 4t. This should be enough for solving that... – ftiaronsem Jan 4 '11 at 12:12
$cos(3t \pi/2): cos(3 \pi/2)=0, cos(3\pi)=-1, cos(9 \pi/2)=0, cos(6 \pi) = 1 \cdots$, but general rule is missing in my mind. – alvoutila Jan 4 '11 at 12:18

## 1 Answer

Well, you nearly got it ;-)

Lets write down the first terms:

$$\begin{array}{cc} k& \quad & \cos \left(\frac{3\pi k}{2} \right) \\ 1& \quad & 0 \\ 2& \quad &-1 \\ 3& \quad & 0 \\ 4& \quad & 1 \\ 5& \quad & 0 \\ 6& \quad &-1 \\ 7& \quad & 0 \\ 8& \qquad & 1 \\ \end{array}$$

So we are obviously searching for something which is $-1$ every second and $1$ every fourth time. Wat comes to mind? $i^k$ Unfortunatelly every first time we have $i$ and every third $-i$

Now we have to find a way to cancl out $i$ every first and third time. We are therefore searching a $x$ so that:

$$\begin{array}{crr} k \quad & i^k & x^k \\ 1 \quad & i &-i\\ 2 \quad & 1 & 1\\ 3 \quad &-i & i\\ 4 \quad &-1 &-1\\ \end{array}$$

Because if we had that, we simply would sum $x^k$ and $i^k$ , divide it by two and would be finished. After a little thinking

$$x^k=(-i)^k$$

comes to mind, since $(-1)^k$ has exactly the alternating properties we are searching.

$$a_n = \frac{i^k+(-i)^k}{2}$$
That's a bit circuitous. Just use the fact that $\cos(x)=(\exp(ix)+\exp(-ix))/2$ and $\exp(3\pi i/2)=-i$ ... – non-expert Jan 4 '11 at 14:31
hahaha, yeah, thats way more direct. Thanks for the comment, I simply hadn't seen $\exp(3\pi i/2)=-i$. But this is very nice... – ftiaronsem Jan 4 '11 at 14:38
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