Re: Maclaurin series of (sin(pi*x))^q
- From: Bart <Reply.In@xxxxxxxxxx>
- Date: Fri, 30 Jun 2006 12:04:50 +0000 (UTC)
On 2006-06-30, Bart <Reply.In@xxxxxxxxxx> wrote:
Note that
sin(t)^q = (exp(i t) - exp(-i t))^q/(2 i)^q
= 2^(-q) (-i)^q sum_{j=0}^q (q choose j) (-1)^(q-j) exp((2j-q) i t)
Thus if q = 2k+1 is odd,
sin(t)^q = (-4)^(-k) sum_{j=0}^k (q choose j) (-1)^j sin((q-2j) t)
Checked with Maple and this formula seems right, but
how exactly do you show this? I assume you write exp((2j-q) i t)
as cos((2j-q)t) + i*sin((2j-q)t) and then some terms cancel. But
I don't quite see how.
OK. After some hand-calculations, I've found out how they cancel and now
I got the picture for q=2k+1 odd.
If q = 2 k is even,
sin(t)^q = (-4)^(-k) (sum_{j=0}^{k-1} (q choose j) (-1)^j cos((q-2j) t)
+ (q choose k) (-1)^k)
Hmmm... checked with Maple and it seems like there's something
wrong...
I'll see if i can check the case q=2k by hand too...
Regards,
Bart
--
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