When I calculate the series expansion of $\log\frac{z}{z-1}$ around $0$ in Mathematica 5.2, I obtain:

 Series[Log[z/(z - 1)], {z, 0, 4}]

 (*Out: (I Pi + Log[z]) + z + z^2/2 + z^3/3 + ... *)

But I can rearrange that expression to the following form:

 Log[z/(z-1)] == Log[z] - Log[(-1) (1-z)] == Log[z] - Log[-1] - Log[1-z]

I then obtain the Taylor expansion of $\log(1-z)$ around $z=0$,

 Log[1-z] = -z - z^2/2 - z^3/3 - ...

and plug it in the above expression, to obtain:

 Log[z/(z-1)] == Log[z] - I Pi + z + z^2/2 + z^3/3 + ...

This last expression differs by $2i\pi$ from what Mathematica gives.

Actually, if I directly evaluate the expansion of Log[z/(z-1)], the latter result is obtained.

  • Can anybody explain to me why?
  • If both are right with appropriate choice of branch cut, etc,. which one should I use in practice then?
  • 1
    $\begingroup$ Yes, I think M should have the sign function there. I verified with Maple, Maple gives this: series(log(z/(z - 1)), z=0); ln(z)+I*csgn(I*z)*Pi+z+(1/2)*z^2+(1/3)*z^3+(1/4)*z^4+(1/5)*z^5+.... notice the I*csgn(I*z) there, which is explained more here $\endgroup$ – Nasser May 18 '15 at 5:59
  • $\begingroup$ The OP's result is not local to V5. In V10.1 Series gives the same result. $\endgroup$ – m_goldberg May 18 '15 at 6:35
  • 1
    $\begingroup$ Your "but I can rearrange" looks iffy to me $\endgroup$ – J. M. is away May 18 '15 at 7:01
  • $\begingroup$ I know it must be an Arg issue, which common users don't care too much in practice. But if we just use the literal expression given by Mathematica, then we will get the wrong(?) answer (differing by 2 I PI). On the other hand, even there is the Arg complication, should Mathematica give the regular expression (say principal value type of consideration)? $\endgroup$ – bsmile May 18 '15 at 20:24

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