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I am trying to solve the following equation, and I can't simplify it even further. Is there any approximation or solution to this equation? $$\psi(r)+r\psi'(r)=\log r + 1$$ where $\psi$ is the Digamma function and $\psi'$is its derivative, and $r$ is a postive integer.

Nooob
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  • I don't know what you mean by "approximation or solution" if you are also asserting that $\psi$ is a digamma function. But it's not hard to solve the give equation. First, the associated homogeneous equation is $r\psi'= \psi$ which we can separate as $\frac{d\psi}{\psi}= \ – user247327 Jun 22 '18 at 19:35
  • By "solving", you mean finding all $r$ satisfying this equation? – Hamed Jun 22 '18 at 19:42
  • @user247327, What?! – Nooob Jun 22 '18 at 19:43
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    $\psi(r)+r \psi'(r)-\ln r-1$ seems to be $>0$ for all $r > 0$, so no real solution. – gammatester Jun 22 '18 at 19:43
  • @Hamed, Yes, finding all $r$ satisfying the equation. – Nooob Jun 22 '18 at 19:44
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    Sorry- I hate the rule that "you can only edit for five minutes". I can never do it that fast! We can write $r\psi'+ \psi= 0$ as $\frac{d\psi}{\psi}= -\frac{dr}{r}$. Integrating, $ln(\psi)= -ln(r)+ c$ so that $\psi(r)= \frac{C}{r}$. To find a solution to the entire equation, let $\psi= \frac{u(r)}{r}$ where u is a function of r to be determined. Then $\psi'= \frac{ru'- u}{r^2}= \frac{u'}{r}- \frac{u}{r^2}$ so $r\psi'+ \psi= u'- \frac{u}{r}+ \frac{u}{r}= u'= ln(r)+ 1$. Integrate that to find u. That does NOT look like a "digamma"! – user247327 Jun 22 '18 at 19:53
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    @user247327: This is no differential equation, a solution $r$ is to be found. E.g. if the 1 is changed to 2, a solution is $r\approx 0.103088$. – gammatester Jun 22 '18 at 19:56
  • @gammatester, using matlab, the function is verified for large $r$. Btw $r$ is a positive integer. – Nooob Jun 22 '18 at 20:00
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    What does this mean "function is verified for large r"? For integer $r$ your function is $0<f(r) < 0.068$ – gammatester Jun 22 '18 at 20:05
  • @gammatester, I calculated the difference betwwen the two terms for $r=1$ to $100$, and as $r$ increases, the difference tends to $0$. – Nooob Jun 22 '18 at 20:07
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    Yes, but I am pretty sure that it does not have a zero, e.g. for $r=100000$ it is $0.8333333333 \cdot 10^{-11}$. For large $r$ values, there should be some asmyptotic formulas for the polygamma functions. Using Maple I get $$f(r) \sim \frac{1}{12r^2}-\frac{1}{40r^4}+O(r^{-6})$$ – gammatester Jun 22 '18 at 20:14
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    You can rewrite your equation as $$ \frac{d}{dr}\Big [r(\psi(r) - \log r)\Big]=0 $$ at the same time for asymptotically large values one has $\psi(x)=\log x -1/2x-1/12x^2+O(x^{-4})$. You can see that alothough your function tends to zero, it actually never is zero. – Hamed Jun 22 '18 at 20:20
  • @Hamed, Thank you very much. – Nooob Jun 22 '18 at 20:40

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