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While trying to improve this interesting answer by @Anastasiya-Romanova I noticed that $$\psi^{(1)}\left(\frac 56\right)-\psi^{(1)}\left(\frac 16 \right)=5\left(\psi^{(1)}\left(\frac 23\right)-\psi^{(1)}\left(\frac 13\right)\right),$$

where $\psi^{(1)}$ is the trigamma function. I don't know how to prove it. Can anybody help us to make her answer complete with the proof of this identity?

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  • $\begingroup$ Using the reflection formula this is equivalent to $\displaystyle 4\pi^2-2\psi^{(1)}\left(\frac {1}{6} \right)=5\left(\frac{4\pi^2}{3}-2\psi^{(1)}\left(\frac {1}{3}\right)\right)$... $\endgroup$ – Shakespeare Oct 1 '14 at 13:57
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It follows from the duplication formula, $$\psi_1(z)+\psi_1\left(z+\frac12\right)=4\psi_1(2z)$$ Plug in $z=1/6, 1/3$ and get $$\psi_1\left(\frac16\right)=4\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)\\\psi_1\left(\frac56\right)=4\psi_1\left(\frac23\right)-\psi_1\left(\frac13\right)$$ Subtract the second equation by the first one and we have the desired result. $$\psi_1\left(\frac56\right)-\psi_1\left(\frac16\right)=5\left(\psi_1\left(\frac23\right)-\psi_1\left(\frac13\right)\right)$$

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