What is the general term $(a_n)$ of the alternating sequence $\displaystyle \cos \left( \frac{3n \pi}{2} \right)$ from $1$ to $\infty$, $n \in \mathbb{N}$ ?
What is the general term $(a_n)$ of the alternating sequence $\cos(3n \pi/2)$?
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0$cos(3t \pi/2): cos(3 \pi/2)=0, cos(3\pi)=-1, cos(9 \pi/2)=0, cos(6 \pi) = 1 \cdots$, but general rule is missing in my mind. – 2011-01-04
1 Answers
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.
So your general term is
$a_n = \frac{i^k+(-i)^k}{2}$
P.S. If anyone knows a tabular environment for MathJax please leave a comment ;-)
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0I didn't get the point. I know that cos(3\pi t/2)=(exp(i 3\pi t/2)+exp(−i3\pi t/2))/2, but why you have to consider exp(3πi/2)=−i, because shouldn't it be rather exp(i 3\pi t/2)=? ? – 2011-01-05