sin((2n+1)x/4)

By Anonymous on Wednesday, May 30, 2001 - 06:32 pm :

Find this sum:
sin(x/4) + sin(3x/4) + sin(5x/4) +...+ sin((2n+1)x/4)
Thank you.
Antonio.

By Oliver Samson (P3202) on Wednesday, May 30, 2001 - 07:14 pm :

Do you mean

π

instead of

x

? If you do, then the answer will cycle between

(1 / 2)^{1 / 2}

2^{1 / 2}

(1 / 2)^{1 / 2}

and 0, depending on the remainder left when

n

is divided by 4.

By Kerwin Hui (Kwkh2) on Thursday, May 31, 2001 - 03:24 pm :

We can use the identity

$\sin y = ℑ e^{i y}$

and get

$\sum_{r = 0}^{n} \sin ((2 r + 1) x / 4) = ℑ \sum_{r = 0}^{n} e^{i (2 r + 1) x / 4}$

$= ℑ [(e^{i x / 4} - e^{i (2 r + 3) x / 4}) / (1 - e^{i x / 2})]$

$= ℑ (1 - e^{i (n + 1) x / 2}) / (e^{- i x / 4} - e^{i x / 4})$

$= ℑ (1 - e^{i (n + 1) x / 2}) / [2 i \sin (- x / 4)]$

$= ℜ (1 / e^{i (n + 1) x / 2}) / [2 \sin (x / 4)]$

$= [1 - \cos ((n + 1) x / 2)] / [2 \sin (x / 4)]$

[An alternative method is to multiply the series by $\sin (x / 4)$ and use the product-to-sum formula. Details omitted]

Kerwin