How can I solve for $a$ in $\frac {k-\frac 32}a+\frac 1{2a^2}+\digamma(a)-\digamma(a+n)=0$, where $\digamma$ is the digamma function?

Asked Jan 03 '19 at 03:19

Active Jan 03 '19 at 05:35

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How can I solve for $a$ for this following equation?

$$\frac {k-\frac 32}a+\frac 1{2a^2}+\digamma(a)-\digamma(a+n)=0,$$

where $\digamma$ represents the digamma function, i.e. it is defined as the logarithmic derivative of the gamma function.

edited Jan 03 '19 at 05:35

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asked Jan 03 '19 at 03:19

userFarkill

1

That's not an equation... – Vegeta the Prince of Saiyans Jan 03 '19 at 03:21
forget to include a =0 – userFarkill Jan 03 '19 at 03:24
What is $k$? A constant? – Kal S. Jan 03 '19 at 05:44
yes. k and n are just a constant – userFarkill Jan 03 '19 at 05:56
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=\frac{1}{a}+\dots + \frac{1}{a+n-1}$. – Kal S. Jan 03 '19 at 06:32
yeap. already tried that. And i'm stuck at the part on how to simplify or change 1a+⋯+1/a+n−1 to a closed form. – userFarkill Jan 06 '19 at 09:24

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