Question

How do I write this in summation notation?

\[(\sin x) (\cos x)^{n - 1} + (\sin x) (\cos x)^{\frac{n - 1}{2}} + (\sin x) (\cos x)^{\frac{\frac{n - 1}{2}}{2}} + (\sin x) (\cos x)^{\frac{\frac{\frac{n - 1}{2}}{2}}{2}} ... \]

n is an odd number and the series ends when (n-1)/2^k = 1, and the last term ends up being sin(x) only (no cos factor). For example, if n was 5, then 4 / 2^2 would be the last term in the series, so it looks like cos^1(x)sin(x) would be the last term, but the last term is actually sin(x) only. So if n was 5, the series would look like:

\[(\sin x) (\cos x)^{4} + (\sin x) (\cos x)^{2} + (\sin x)\]

Answer 1

i am just trying.
the cos powers are 1,2,4,8,16.... so, 2k
starts from 0 and ends to n-1 and factoring out sin x,
\(\large \sum \limits_{k=0}^{n-1} \sin x \cos x^{2k}\)

Answer 2

\(\large \sum \limits_{k=0}^{n-1} \sin x (\cos x)^{2k}\)

Answer 3

If n = 5, then for your term you have cos^8(x)

Answer 4

ohh...

Answer 5

how about
\(\large \sum \limits_{k=0}^{(n-1)/2} \sin x (\cos x)^{2k}\)

Answer 6

That one would work, I guess. I'll post another problem

Answer 7

okk..

Answer 8

I got something slightly different. I used n = 2^m + 1 and set the sum over m

Answer 9

i think we need to add restriction that k should be odd..

Answer 10

and n as a power ?
you'll get odd powers...
m=1, n=3
m=2, n=5

Answer 11

like if n = 5, then m = 2

Answer 12

how about this :
\(\large \sum \limits_{k=0}^{(n-1)/2} \sin x (\cos x)^{2^k}\)

Answer 13

Oh

Answer 14

2k will work, 2^k will not work..

Answer 15

You are right