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Deriving Arctan(z) formula using series.

  1. Sep 19, 2012 #1
    Hi, I am still trying to learn latex so this might fail but anyway, I am trying to establish the identity

    [itex]arctan(z) = \frac{1}{2i}log(\frac{1+iz}{1-iz})[/itex]

    By using the power series of

    [itex]log(1+iz)[/itex] and [itex]log(1-iz)[/itex]

    I have found the power series for both of these functions as well as arctan(z)

    [itex]arctan(z) = \Big( \sum_{n=0}^\infty\frac{(-1)^n(z^(2n+1))}{2n+1} \Big) [/itex]
    [itex]log(1+iz) = -\Big( \sum_{n=0}^\infty\frac{(-1)^(n+1)((iz)^(n+1))}{n_1} \Big) [/itex]
    [itex]log(1-iz) = -\Big( \sum_{n=0}^\infty\frac{(-1)^(n+1)((-iz)^(n+1))}{n_1} \Big) [/itex]

    I then computed

    [itex]\frac{1}{2i}(log(1+iz)-log(1-iz))[/itex]

    I get a sum which doesn't really look anything like that expected. I have tried simplifying but still can't deduce that they are equal. I thought I could perhaps differentiate both sums n times and show that these are equal but that also gets very very messy.. can somebody point me in the right direction?
     
  2. jcsd
  3. Sep 19, 2012 #2

    Do you have to use series? Because you can as well show that the derivatives in both sides of the equality you're

    trying to prove are identical (after determining the branch of the complex logarithm, say with argument equal to zero for

    positive reals), and since both sides equal zero when [itex]\,z=0\,[/itex] you're done.

    DonAntonio
     
  4. Sep 19, 2012 #3
    I should not have posted this question here, I am trying to move it at the moment because it is homework.

    But yes I am asked to use series. I know how to derive it using the exponentials of sin & cos.

    I read that if I can show that for some c in U in which both my 2 series I am comparing are defined in U, that [itex]f^n (c)=g^n (c) for all n, then f(z)=g(z)[/itex], is this what you're saying? that they are hopefully equal at [itex]z=0[/itex]
     
  5. Sep 19, 2012 #4


    No, what I am saying is way simpler and basic. If two functions [itex]\,f,g\,[/itex] derivable in some open interval (or some open ball in complex plane) I are s.t.

    $$f'(x)=g'(x)\,\,,\,\,\forall\,x\in I$$

    then [itex]\,\forall \,x\in I\,\,,\,\,f(x)=g(x)+C\,\,,\,C=\,\text{a constant}[/itex]

    DonAntonio
     
  6. Sep 19, 2012 #5
    Okay thanks, I do understand that but like I said, even as they are, I can't manipulate my two series to look identical already, so even if I differentiate them I will just end up with the same problem.. :(
     
  7. Sep 19, 2012 #6
    Ok this will take me a while but I will show my working and what I CAN deduce..

    [itex]\frac{1}{2i}log(\frac{1+iz}{1-iz}) = \frac{1}{2i}(\Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}(iz)^{n+1}}{n+1} \Big) + \Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}(-iz)^{n+1}}{n+1} \Big)) [/itex]

    [itex]\frac{1}{2i}log(\frac{1+iz}{1-iz}) = \frac{1}{2i}\Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}((-iz)^{n+1}-(iz)^{n+1}}{n+1} \Big) [/itex]

    [itex]\frac{1}{2i}log(\frac{1+iz}{1-iz}) = \frac{1}{2i}\Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}(iz)^{n+1}((-1)^{n+1}-1)}{n+1} \Big) [/itex]

    [itex]\frac{1}{2i}log(\frac{1+iz}{1-iz}) = \Big( \sum_{n=0}^\infty\frac{(-1)^{n+1}i^{n}z^{n+1}((-1)^{n+1}-1}{2(n+1)} \Big) [/itex]

    And from this point on I am completely screwed, the power series for [itex]arctan(z)[/itex] doesn't even contain an [itex]i[/itex] and that is only the start of my problems...
     
    Last edited: Sep 19, 2012
  8. Sep 19, 2012 #7


    Don't you know the derivatives of the functions arctangent z , log z and/or the chain rule?! I guess you

    surely didn't mean this as then how could you work with power series...?

    $$(\arctan z)'=\frac{1}{1+z^2}$$

    $$\left[\frac{1}{2i}\log\left(\frac{1+iz}{1-iz}\right)\right]'=\frac{1}{2i}\frac{1-iz}{1+iz}\frac{2i}{(1-iz)^2}=\frac{1}{(1+iz)(1-iz)}=\frac{1}{1+z^2}$$

    DonAntonio
     
  9. Sep 19, 2012 #8
    Oh no I do sorry, I think I misinterpreted what you were saying. My only problem is that the question explicitly tells me to obtain the power series for the logs about z=0 and use them to establish the identity. I will stare at what you wrote for a while and see if I can put that to use using the questions "criteria".

    Ok yes, If I write the (1/2i)Log((1+iz)/(1-iz)) in terms of the power series and differentiate termwise, I should be able to get to the required answer, I'll give it a go!
     
  10. Sep 19, 2012 #9
    Still ending up with two series that I cannot show are equal. I completely understand everything you have said but the only problem is that I need to use the power series of the logs!...

    I just essentially tried what you said but instead of directly differentiating (1/2i)Log((1+iz)/(1-iz)) I wrote it as (1/2i)(log(1+iz)-log(1-iz)) and used the power series and differentiated term wise. I end up with an ugly series.
     
  11. Sep 19, 2012 #10
    OK I FINALLY GOT IT, I'VE BEEN WORKING ON THIS FOR 2 DAYS!

    I essentially used your method which I did actually try in a much longer way yesterday. The problem was that my lecture notes say that you can manipulate

    [itex]\frac{1}{1+z^2}[/itex]

    Into

    [itex]\frac{1}{1-(-z^2)}[/itex]

    And thus

    [itex]\frac{1}{1+z^2} = \Big( \sum_{n=0}^\infty(-z^2)^n \Big) = \Big( \sum_{n=0}^\infty(-1)^{n}z^{2n} \Big)[/itex]

    When according to wolframalpha this is in fact not true at all!!! and that

    [itex]\frac{1}{1+z^2} = \Big( \sum_{n=0}^\infty\frac{1}{2}z^n((-i)^n+i^n)\Big)[/itex]

    I hate my bloody lecturer.. What the hell is going on here? I was under the impression that this could be done, in fact I did it in my derivation for the power series of [itex]arctan(z)[/itex] and arrived at the correct answer, but using the simple

    [itex]\Big( \sum_{n=0}^\infty(-1)^{n}z^{2n} \Big)[/itex]

    And equating it to the derivative of

    [itex]\frac{1}{2i}log(\frac{1+iz}{1-iz})[/itex]

    Doesn't seem to work at all.

    Could somebody please explain what is going on? Can [itex]\frac{1}{1+z^2}[/itex] be manipulated in the aforementioned way or not? Why not?
     
    Last edited: Sep 19, 2012
  12. Sep 19, 2012 #11
    It's just the sum of an infinite geometric series:
    [tex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n[/tex]
    Now let
    [tex]x = -z^2[/tex]

    For the rest of the problem, note that
    [tex]\log \left( \frac{1+iz}{1-iz} \right) = \log(1+iz) - \log(1-iz)[/tex]
    and apply the series
    [tex]\log(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}[/tex]

    I haven't worked it all out, but I think that approach should work.
     
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