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Mary Boas

  1. Mar 15, 2015 #1
    On page 671 Mary Boas has her Theorem III for that chapter. Roughly it tells us that if f(z) -a complex function- is analytic in a region, inside that region f(z) has derivatives of all orders. We can also expand this function in a taylor series.

    I get the part about a Taylor series, that's pretty straightforward. But what does she mean about this region and the derivatives. I get that there are higher order derivatives. But I wish she would have given an example of one such function. Also, isn't analytic here only tested for by taking the first derivative? So if we do that we can't just assume that there will be higher order derivatives up to the nth order.
     
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  3. Mar 15, 2015 #2

    wabbit

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    Ah but that is precisely the great miracle of complex differentiation ! I do not have that book but I am surprised she doesn't introduce this before. And if she doesn't, the first few pages of any introductory book on complex analysis should cover that - or have a look at https://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions
     
  4. Mar 15, 2015 #3
    I'm still confused. Say for example you have d/dx z^2 which gives by the equivalent power rule 2z, I assume the next derivative gives 2, then next is 0. So how does this satisfy the analytic for all higher derivatives theorem here? Thanks for your time and help by the way. :D
     
  5. Mar 15, 2015 #4

    wabbit

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    I am not sure I get your question here, it sounds as if you're saying that the higher derivatives of z2 are not immediately obvious to you. Can you clarify ?

    Because if that's the case, then you're not ready to take a course in complex analysis at all. What is your background in maths, what courses have you completed ?

    However if you were able to study and understand the first 670 pages of that book with no issue, then there's obviously something I'm missing here... So I'll stop speculating and will wait for your clarification:)
     
    Last edited: Mar 15, 2015
  6. Mar 15, 2015 #5
    LOL yes there's been a misunderstanding. I've done multvar calc, ODEs and a few other things. I'm saying that since

    d/dx z2 = 2z
    d2/dx2=2
    d3/dx3=0=dn>3/dn>3

    And z2 is analytic then do we continue to call it analytic even though any derivative higher than the third is still just 0? If that's the case, what makes it "not" have a derivative and is not analytic? Is it simply when the derivative is such that plugging in for a value in the region that this complex function is being analyzed leads to an undefined result?

    At first I was thinking well if it's going to have a derivitive by this definition we must be talking about functions like Sinz or ez but then there's the polynomial one we've discussed and it qualifies under this definition. So I'm uncertain if I get what the definition is telling me. I think perhaps it's the fact that some derivitives "explode" for certain values plugged into them, such as when your denominator gets turned into zero upon plugging in for values in a region. Is this correct?

    BTW, your last response made me laugh man. LOL
     
  7. Mar 16, 2015 #6

    wabbit

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    OK then -
    (just one detail it bugs me to see "d/dx(z2)", this should be d/dz)
    Yes z2, or any polynomial for that matter, is analytic - no proof really needed, it's just an immediate application of the definition. These are in a way the "most analytic" functions one can think of. I don't understand why you think 0 is not a valid derivative, but it is.
     
  8. Mar 16, 2015 #7
    I wasn't saying it's not valid, though I see how I kinda did say it. LOL But I talked with a math friend and got the issue figured out. It seems it's an issue of the region that is in question. :D Thank you so much for your help though. It was a tough one to explain what my confusion was. We would have been chatting for weeks. LOL Again thank you though. :D
     
  9. Mar 16, 2015 #8
    LOL I also just noticed what you were saying about the d/dx thing. Yeah, my mistake. I will write d/dz z2 a hundred times tonight as my punishment. I'm silly sometimes:rolleyes:.
     
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