Solved: "Prove that cos⁡(πz4)−sin⁡(πz4)=∏n=1∞(1+(−1)nz2n−1)"

Ahmed Stewart

Answered question

2022-01-27

Prove that

\cos (\frac{π z}{4}) - \sin (\frac{π z}{4}) = \prod_{n = 1}^{\infty} (1 + \frac{{(- 1)}^{n} z}{2 n - 1})

Natalia Thomas

Beginner2022-01-28Added 7 answers

\prod_{n = 1}^{\infty} (1 + \frac{{(- 1)}^{n} z}{2 n - 1}) = \prod_{n = 0}^{\infty} (1 - \frac{z}{4 n + 1}) \prod_{n = 1}^{\infty} (1 + \frac{z}{4 n - 1}) =

= (1 - z) \prod_{n = 1}^{\infty} (\frac{4 n (1 - \frac{z - 1}{4 n})}{4 n (1 + \frac{1}{4 n})}) \prod_{n = 1}^{\infty} (\frac{4 n (1 + \frac{z - 1}{4 n})}{4 n (1 - \frac{1}{4 n})}) = (1 - z) \frac{\prod_{n = 1}^{\infty} (1 - {(\frac{z - 1}{4 n})}^{2})}{\prod_{n = 1}^{\infty} (1 - {(\frac{1}{4 n})}^{2})}

= (1 - z) \frac{\sin (\frac{z - 1}{4} π)}{\frac{z - 1}{4} π} \frac{\frac{1}{4} π}{\sin (\frac{1}{4} π)} = - \sqrt{2} \sin (\frac{π z - π}{4}) = - 2 \sin (\frac{π}{4}) \sin (\frac{π z - π}{4})

= \cos (\frac{π z - π}{4} + \frac{π}{4}) - \cos (\frac{π z - π}{4} - \frac{π}{4}) = \cos (\frac{π z}{4}) - \sin (\frac{π z}{4})

because of

\sin x \sin y = - \frac{1}{2} (\cos (x + y) - \cos (x - y))

Note:
The decisive step was

\frac{1}{\sin (\frac{1}{4} π)} = 2 \sin (\frac{π}{4})

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