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In this brief tutorial we evaluate the Trigamma, Tetragamma, and other higher order Polygamma functions at small rational arguments:$$(01) \quad \psi_{m \ge 1}(z) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+z)^{m+1}}$$We will have frequent need of the reflection formula, which is obtained by repeated differentiation of the reflection formula for the Digamma function:$$(02) \quad \psi_0(z) - \psi_0(1-z) = -\pi\cot \pi z \, \Rightarrow$$$$(03) \quad \psi_{m \ge 1}(z) + (-1)^{m+1}\psi_{m \ge 1}(1-z) = -\frac{d^m}{dz^m}\, \pi\cot \pi z$$From the definition (01) above, we have the trivial case:$$(04) \quad \psi_{m\ge 1}(1) = (-1)^{m+1}m!\, \zeta(m+1)$$. Hence$$\psi_1(1) = \zeta(2)$$$$\psi_2(1) = -2\, \zeta(3)$$$$\psi_3(1) = 6\, \zeta(4)$$$$\psi_4(1) = -24\, \zeta(5)$$$$\psi_5(1) = 120\, \zeta(6)$$$$\psi_6(1) = -720\, \zeta(7)$$
The case for $$z=1/2$$ is equally straightforward. Setting $$z=1/2$$ in (01) above we get:$$\psi_{m \ge 1} \left(\tfrac{1}{2}\right) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+1/2)^{m+1}}=
(-1)^{m+1}2^{m+1}\,m!\, \sum_{k=0}^{\infty} \frac{1}{(2k+1)^{m+1}} \equiv$$$$(-1)^{m+1}2^{m+1}\,m!\, \left[ \sum_{k=1}^{\infty} \frac{1}{k^{m+1}} - \frac{1}{2^{m+1}}\, \sum_{k=1}^{\infty} \frac{1}{k^{m+1}} \right]=$$$$(-1)^{m+1}2^{m+1}\,m!\, \left[ \left(1-\frac{1}{2^{m+1}} \right) \, \zeta(m+1) \right]$$
Hence$$(05) \quad \psi_{m\ge 1}\left(\tfrac{1}{2}\right) = (-1)^{m+1}(2^{m+1}-1)\,m!\, \zeta(m+1)$$. and$$\psi_1\left(\tfrac{1}{2}\right) = 3\zeta(2)$$$$\psi_2\left(\tfrac{1}{2}\right) = -14\, \zeta(3)$$$$\psi_3\left(\tfrac{1}{2}\right) = 90\, \zeta(4)$$$$\psi_4\left(\tfrac{1}{2}\right) = -744\, \zeta(5)$$$$\psi_5\left(\tfrac{1}{2}\right) = 7560\, \zeta(6)$$$$\psi_6\left(\tfrac{1}{2}\right) = -91440\, \zeta(7)$$
Comments and/or questions should be posted here:
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The case for $$z=1/2$$ is equally straightforward. Setting $$z=1/2$$ in (01) above we get:$$\psi_{m \ge 1} \left(\tfrac{1}{2}\right) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+1/2)^{m+1}}=
(-1)^{m+1}2^{m+1}\,m!\, \sum_{k=0}^{\infty} \frac{1}{(2k+1)^{m+1}} \equiv$$$$(-1)^{m+1}2^{m+1}\,m!\, \left[ \sum_{k=1}^{\infty} \frac{1}{k^{m+1}} - \frac{1}{2^{m+1}}\, \sum_{k=1}^{\infty} \frac{1}{k^{m+1}} \right]=$$$$(-1)^{m+1}2^{m+1}\,m!\, \left[ \left(1-\frac{1}{2^{m+1}} \right) \, \zeta(m+1) \right]$$
Hence$$(05) \quad \psi_{m\ge 1}\left(\tfrac{1}{2}\right) = (-1)^{m+1}(2^{m+1}-1)\,m!\, \zeta(m+1)$$. and$$\psi_1\left(\tfrac{1}{2}\right) = 3\zeta(2)$$$$\psi_2\left(\tfrac{1}{2}\right) = -14\, \zeta(3)$$$$\psi_3\left(\tfrac{1}{2}\right) = 90\, \zeta(4)$$$$\psi_4\left(\tfrac{1}{2}\right) = -744\, \zeta(5)$$$$\psi_5\left(\tfrac{1}{2}\right) = 7560\, \zeta(6)$$$$\psi_6\left(\tfrac{1}{2}\right) = -91440\, \zeta(7)$$
Comments and/or questions should be posted here:
http://mathhelpboards.com/commentary-threads-53/commentary-evaluations-higher-order-polygamma-functions-8174.html
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