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I Validity of laws of logs

  1. Oct 15, 2017 #1

    CAF123

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    Perhaps this is a simple question but I can't see a way to understand it. I have some expression of the form ##\ln(1-1/(1-x))## and I perform the following manipulations using simple algebra and laws of logs:

    $$\ln \left(1-\frac{1}{1-x}\right) = \ln \left(\frac{1-x-1}{1-x}\right) = \ln \left(\frac{x}{(x-1)}\right) = -\ln \left(\frac{x-1}{x}\right) = -\ln \left(1-\frac{1}{x}\right)$$

    I am doing such manipulations under the premise of ##x \in (0,1)##. However, if I put x equal to 1/2 then the left hand side does not match the right hand side. It does for x otherwise though. So, is it correct that the above rewriting is only valid for x not 1/2 and ,if so, why and how to rewrite ##\ln(1-1/(1-x))## in terms of ##\ln(1-1/x)## valid for ##x \in (0,1)##?
     
    Last edited: Oct 15, 2017
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  3. Oct 15, 2017 #2

    Stephen Tashi

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    If we're dealing with the real number system, when ##x = 1/2## then ##\ln \left(1-\frac{1}{1-x}\right) ## doesn't exist because -1 isn't in the domain of ##ln()##.
     
  4. Oct 15, 2017 #3

    CAF123

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    I think at the end of the day I want to analytically continue the x. The law of logs ##b\ln(a) = \ln(a^b)## is only valid if a is positive... I rewrote my term as follows $$\ln(1-\frac{1}{1-x}) = \ln(\frac{x}{x-1}) = \ln(\frac{-x}{x-1}) + i\pi = -\ln(\frac{1}{x}-1) + i\pi.$$ Then used ##\ln(-z) = \ln z + i\pi## again to write $$\ln(1-1/(1-x)) = 2i\pi - \ln(1-1/x)$$
    Is this now a correct rewriting of log(1-1/(1-x)) given that my x is restricted to x in (0,1)?
     
  5. Oct 16, 2017 #4

    Stephen Tashi

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    Are you trying to find an identity that applies to (the principal value of)##ln(z)## for ##z = a + bi## where ##b## is an arbitrary real number? Or are you only trying to find an identity that works just for the case ##b = 0##?
     
  6. Oct 16, 2017 #5

    Erland

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    Let's change it a little, to get rid of the problem with analytic continuation and log of negative numbers:

    Assume ##x\in (0,1)##. Then:

    $$\ln \left(\frac{1}{1-x}-1\right) = \ln \left(\frac{1-(1-x)}{1-x}\right) = \ln \left(\frac{x}{1-x}\right) = -\ln \left(\frac{1-x}{x}\right) = -\ln \left(\frac{1}{x}-1\right)$$

    This holds also for ##x=\frac12##, because then both sides are ##0##.
     
  7. Oct 18, 2017 #6

    CAF123

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    Hi @Stephen Tashi,
    Upon further analysis of my problem, I see that I want my x to take values in (-1,1) and not (0,1). I want to get an identity involving logs with b=0 only.

    I'm given that the analytic continuation of terms of the form ##\ln(1+1/x_-)## and ##\ln(1-1/x_+)## are given by replacements ##x_+ \rightarrow x - i\theta(x)\epsilon## and ##x_- \rightarrow x + i \theta(-x)\epsilon## and taking the limit ##\epsilon \rightarrow 0^+##. I would like to consider what these terms look in each of the regions ##x>0## and ##x<0##. In the latter case, the above formulae tell me that ##x_+ = x## and ##x_- = x + i\epsilon##. So, in this region, I can simply write ##\ln(1-1/x_+) = \ln(1-1/x)##. The other term ##\ln(1+1/x_-)## is more difficult:

    The analytic continuation tells me that in region x<0, ##x_- ## is at ##x+i\epsilon##, so at angle ##\pi-\epsilon## from the negative real axis. Therefore I believe I can use the formula $$\ln(-z) = \ln(e^{i\pi}z) = i\pi + \ln z$$ but what is my ##z## here?

    Thanks for any comments!
     
    Last edited: Oct 18, 2017
  8. Oct 18, 2017 #7

    Stephen Tashi

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    As I understand "analytic continuation" it begins with an analytic function defined on a open subset of the complex plane (rather than with "terms" of some form). So what function and what open subset of the complex plane do you wish to consider?
     
  9. Oct 19, 2017 #8

    Erland

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    I think all this with analytic continuation and principal values misses the point. It was unfortunate that CAF123 chose an argument of ln which was negative, and got problems because log is multi-valued. But this has little to do with the original problem with an expression u such that u = -u (for x=1/2). But this is no problem if u=0, which is easily seen to be the case if we change the original expressions a little, so that we only get ln of positive numbers, which I did my post #5 above.
     
  10. Oct 19, 2017 #9

    CAF123

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    Ah sorry I guess I meant to say the ##\ln( 1 \pm 1/x_{\mp})## are the functions I wish to consider. Perhaps it's worth explaining a little more what I wish to do. Basically I've done an exercise and my result (a function involving sums of logs and square of logs) is only a function of this x. I'm comparing to a result which involves this x (only in rational prefactors which we probably don't need to worry about) but also ##x_+## and ##x_-## through the logs. So I wish to rewrite this latter result entirely in terms of x so I can compare to my result. To do so, I have to use the proposed 'analytic continuations' ##x_+ \rightarrow x - i\theta(x)\epsilon## and ##x_- \rightarrow x + i\theta(-x)\epsilon##. I'm not sure if 'analytic continuation' is the correct way of saying what I wish to do but I just wish to rewrite e.g ##\ln(1 + 1/x_-)## solely in terms of x.

    I try to see if the results match in a certain region of x. E.g for x<0, ##x_+ \rightarrow x## and ##x_- \rightarrow x + i\epsilon##. I can now insert this ##x_-## into ##\ln(1 + 1/x_-)## but it's how to deal with the ##i\epsilon## thats confusing me a bit.

    My thinking is that these ##x_+## and ##x_-## are introduced as a compact way of writing the log to cover both regions, x>0 and x<0.

    Is my intentions a little clearer? Thanks!
     
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