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Kevin Casto
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I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the duplicationmultiplication formula is simpler;, andsimpler: we have $$\Pi\left(\frac{z}{m}\right) \, \Pi\left(\frac{z-1}{m}\right) \cdots \Pi\left(\frac{z-m+1}{m}\right) = \left(\frac{(2 \pi)^m}{2 \pi m}\right)^{1/2} \, m^{-z} \, \Pi(z)$$ instead of $$\Gamma\left(\frac{z}{m}\right) \, \Gamma\left(\frac{z-1}{m}\right) \cdots \Gamma\left(\frac{z-m+1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - z} \; \Gamma(z);$$

the infinite product definitions reduce from $$\begin{align} \Gamma(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}} \\ \Gamma(z) &= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \\ \end{align}$$ to $$\begin{align} \Pi(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{(z+1)\cdots(z+n)} = \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}} \\ \Pi(z) &= e^{-\gamma z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}; \\ \end{align}$$ and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$$$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the duplication formula is simpler;, and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the multiplication formula is simpler: we have $$\Pi\left(\frac{z}{m}\right) \, \Pi\left(\frac{z-1}{m}\right) \cdots \Pi\left(\frac{z-m+1}{m}\right) = \left(\frac{(2 \pi)^m}{2 \pi m}\right)^{1/2} \, m^{-z} \, \Pi(z)$$ instead of $$\Gamma\left(\frac{z}{m}\right) \, \Gamma\left(\frac{z-1}{m}\right) \cdots \Gamma\left(\frac{z-m+1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - z} \; \Gamma(z);$$

the infinite product definitions reduce from $$\begin{align} \Gamma(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}} \\ \Gamma(z) &= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \\ \end{align}$$ to $$\begin{align} \Pi(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{(z+1)\cdots(z+n)} = \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}} \\ \Pi(z) &= e^{-\gamma z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}; \\ \end{align}$$ and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

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Yemon Choi
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I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}$,$$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the duplication formula is simplersimpler;, and the Riemann zeta functional equation reduces from $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s)$.$$2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumfrencecircumference/diameter instead of the much more natural circumfrencecircumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$ instead of $\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}$, the duplication formula is simpler, and the Riemann zeta functional equation reduces from $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$ to $2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s)$.

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumfrence/diameter instead of the much more natural circumfrence/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the duplication formula is simpler;, and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

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Kevin Casto
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I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$ instead of $\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}$, the duplication formula is simpler, and the Riemann zeta functional equation reduces from $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$ to $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s)$$2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s)$.

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumferencecircumfrence/diameter instead of the much more natural circumferencecircumfrence/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$ instead of $\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}$, the duplication formula is simpler, and the Riemann zeta functional equation reduces from $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$ to $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s)$.

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$ instead of $\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}$, the duplication formula is simpler, and the Riemann zeta functional equation reduces from $\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$ to $2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s)$.

I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumfrence/diameter instead of the much more natural circumfrence/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

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Pete L. Clark
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Yemon Choi
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Kevin Casto
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