Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to derive Euler's infinite product formula

$$\displaystyle \sin(\pi z) = \pi z \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right)$$

by using Euler's reflection equation $\Gamma(z)\Gamma(1-z) \sin(\pi z) = \pi$ and the definition of $\Gamma(z)$ as an infinite product, namely

$$\displaystyle \Gamma(z) := \frac{1}{z} \prod_{k=1}^\infty \frac{(1+\frac{1}{k})^z}{1+\frac{z}{k}}.$$

To be precise, I obtain that

$$\sin(\pi z) = \pi z(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right)$$

hence I wish to prove

$$(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right) = \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right).$$

I multiplied things out and got it to the form

$$(1-z) \prod_{k=1}^\infty \frac{1 + \frac{1}{k} + \frac{z(1-z)}{k^2}}{1 + \frac{1}{k}} = (1-z) \prod_{k=1}^\infty \left( 1 + \frac{\frac{z(1-z)}{k}}{1+\frac{1}{k}}\right)$$ however the $(1-z)$ factor out front is giving me some trouble; I'm not sure how to proceed.

share|improve this question
    
Something looks fishy: Your final product appears to have its zeros in the wrong places. –  Harald Hanche-Olsen Jun 18 '12 at 12:30

2 Answers 2

up vote 8 down vote accepted

Use the fact that $\Gamma (1-z) = -z\, \Gamma(-z)$ and then:

$$\Gamma(1-z)\Gamma(z) = -z \, \Gamma(-z)\Gamma(z) = -z \cdot \frac{1}{-z}\cdot \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{\left(1 + \frac{z}{k} \right)\left(1 - \frac{z}{k} \right) } = \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{1 - \frac{z^2}{k^2}} = \frac{\pi}{\sin \pi z}$$

share|improve this answer

When k is equal to z it will be equal to 1, 1-1=0. A product series with a zero in it is 0. Since sin(z*pi) is always zero, both sides are equal. But I'm not sure what that is supposed to mean. If I did the equation in two parts leaving out z=k, the part before the z, and the part after, I will get a completely different answer than zero. Is this thing a hack?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.