Can Euler's reflection formula be used to calculate $\Gamma (1/3)$? (That is, to express in in some closed form.) When trying to apply it, I am always falling in a loop.
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do you mean $\Gamma(1/3)$ ?? – S L May 02 '14 at 18:12
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Do you mean how it is calculated in practice? – May 02 '14 at 18:40
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3$\Gamma(1/3)$ is a transcendental number algebraically independent of $\pi$; no really explicit formula for this number is known. All that is explained, say, in Wikipedia. What is your question exactly? – Grigory M May 02 '14 at 18:48
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I wonder how is it calculated gamma (1/3) using the formula of Euler reflection and also like to see a demonstration gamma (1/3) = 2.6 ... – user147405 May 02 '14 at 18:58
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2This thread may help. – Raymond Manzoni May 02 '14 at 19:10
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Are you asking about a pencil-and-paper calculation? This business about "falling in a loop" suggests you are applying Euler's reflection formula repeatedly and expecting to get something new from that alone. – hardmath May 02 '14 at 21:31
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I think you are not understanding my question. Would like someone to prove me ajudadesse Gamma (1/3) = 2.6 .... using the formula of reflection. =) – user147405 May 02 '14 at 22:06
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Although Euler's reflection formula is satisfied by the gamma function:
$$ \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}, \;\; \forall x \not\in \mathbb{Z} $$
and this suffices to determine $\pm\;\Gamma(\frac{1}{2})$, it does not define all values of the function $\Gamma(x)$. In particular there are many functions that satisfy the reflection formula above, but which have differing values at $x=\frac{1}{3}$.
One easily sees that setting $x=\frac{1}{2}$ in the reflection formula gives us $\Gamma(\frac{1}{2})^2 = \frac{\pi}{1}$, so $\Gamma(\frac{1}{2}) = \sqrt{\pi}$ if we know this value is positive (clear from the integral definition of the gamma function).
However consider any positive real base $a \gt 0$ and define $f_a(x) = a^{x - \frac{1}{2}}$.
Then $f_a(x) f_a(1-x) = a^0 = 1$, and one easily verifies that $f_a(x) \Gamma(x)$ will satisfy Euler's reflection formula for noninteger $x$. By varying $a$ we can obtain any positive value for $f_a(\frac{1}{3}) \Gamma(\frac{1}{3})$ that we like.
Note also that substituting $-\Gamma(x)$ for $\Gamma(x)$ in the reflection formula leaves the product $\Gamma(x) \Gamma(1-x)$ unchanged, so this is another demonstration that the reflection formula alone does not define the gamma function.
The integral definition of $\Gamma(\frac{1}{3})$ can give precise approximations as the link in Raymond Manzoni's Comment explains, and Grigory M's Comment pointing you to Wikipedia should provide more rapid approximations of this transcendental number.
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