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Math Help - Analytic functions

  1. #1
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    Analytic functions

    I have 2 statements, which I don't know how to show that it is true.

    1. An analytic function is differentiable for any order of derivatives.
    If I prove this, I may differentiate analytic function until it is zero.
    2. A bounded entire function must be a constant function.
    For this case, I have no idea to prove it.
    Thank you.
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  2. #2
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    Both statements are consequences of the Cauchy Integral Formula. For the first, let f be continuous on a contour \Gamma and define

    F(z)=\int_\Gamma\frac{f(\zeta)}{\zeta-z}d\zeta

    for each z not on \Gamma. Then F is analytic at each point not on the contour, and the derivitive is

    F^\prime (z)=\int_\Gamma\frac{f(\zeta)}{(\zeta-z)^2}d\zeta.

    See if you can prove this from Cauchy. Then do some reasoning to conclude that all the derivatives exist for an analytic function.

    The statement of your second question is called Liouville's Theorem, and it is a beautiful theorem of analysis. To prove it, one generally proceeds by letting f be analytic inside and on a circle of radius R centered at z_0. If f has a maximum of M on this circle, then

    |f^{(n)}(z_0)|\leq\frac{n!M}{R^n}

    for n\in\mathbb{N}. This lemma is proved by first using Cauchy's Formula on f, then noticing that the integrand of the resulting integral is bounded by M/R^{n+1}. Then use the standard bound

    \left| \int_\Gamma f(z) dz\right| \leq Ml(\Gamma), where \Gamma is your contour of length l. From this, you should arrive at the equation given by the lemma. Liouville's Theorem follows by taking n=1, and letting R go to infinity. A simple bit of reasoning leads to the conclusion that f is constant.

    Post again if you get stuck on any details.
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  3. #3
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    Quote Originally Posted by noppawit View Post
    I have 2 statements, which I don't know how to show that it is true.

    1. An analytic function is differentiable for any order of derivatives.
    If I prove this, I may differentiate analytic function until it is zero.
    I'm not sure what you mean by the text in bold above, but: The function:

    f(z)=e^z

    is analytic but never zero nor are any of its derivatives.
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    I'm not sure what you mean by the text in bold above, but: The function:

    f(z)=e^z

    is analytic but never zero nor are any of its derivatives.
    Anyway, now I am not sure what I understood is correct or not. So, what you mean is this statement is false, right? And if, it is analytic in some domain, it is not differentiable for any order, right?
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  5. #5
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    Quote Originally Posted by noppawit View Post
    Anyway, now I am not sure what I understood is correct or not. So, what you mean is this statement is false, right? And if, it is analytic in some domain, it is not differentiable for any order, right?
    No analytic implies differentiable for any order, it was the statement in bold that I did not understand and the natural interpretation of which seemed false.

    CB
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  6. #6
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    If we differentiate  e^x n times and n tends to infinity , will the function become zero ?


     e^x = \lim_{n\to\infty} ( 1+ \frac{x}{n})^n

     (e^x)' = \lim_{n\to\infty} \frac{n}{n} ( 1 + \frac{x}{n})^{n-1}

     (e^x)'' = \lim_{n\to\infty} \frac{n(n-1)}{n^2} ( 1 + \frac{x}{n})^{n-2}

    ................
    ................
    ................

     \lim_{n\to\infty} (e^x)^{(n)} = \lim_{n\to\infty} \frac{ n!}{n^n}


    Now , it is a constant , if we then diffentiate it one more time , it will become zero .
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  7. #7
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    Nonsense!
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  8. #8
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    Quote Originally Posted by simplependulum View Post
    If we differentiate  e^x n times and n tends to infinity , will the function become zero ?


     e^x = \lim_{n\to\infty} ( 1+ \frac{x}{n})^n

     (e^x)' = \lim_{n\to\infty} \frac{n}{n} ( 1 + \frac{x}{n})^{n-1}

     (e^x)'' = \lim_{n\to\infty} \frac{n(n-1)}{n^2} ( 1 + \frac{x}{n})^{n-2}

    ................
    ................
    ................

     \lim_{n\to\infty} (e^x)^{(n)} = \lim_{n\to\infty} \frac{ n!}{n^n}


    Now , it is a constant , if we then diffentiate it one more time , it will become zero .
    What HallsOfIvy said.

    \frac{d}{dz} e^z=e^z

    hence:

    \frac{d^n}{dz^n}e^z=e^z

    so now what is:

    \lim_{n\to \infty} \frac{d^n}{dz^n}e^z


    CB
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  9. #9
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    So far, everything made me confuse.
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  10. #10
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    What is upsetting everyone is your statement:
    "If I prove this (that analytic implies infinitely differentiable), I may differentiate analytic function until it is zero."

    It is true that an analytic function is infinitely differentiable but what do you mean by "until it is zero"?

    You may be thinking of polynomials. All polynomials are infinitely differentiable and every time you differentiate you reduce the degree by one so if you differentiate a polynomial of degree n n-times it reduces to a constant and then the n+1 derivative (and all succeeding derivatives) are identically 0.

    But that is not true for general analytic functions. The function f(z)= e^z is analytic and so infinitely differentiable and all derivatives are e^z. Also y= cos(z) and sin(z) are analytic and all derivatives are one of cos(z), sin(z), -cos(z), -sin(z).
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    What is upsetting everyone is your statement:
    "If I prove this (that analytic implies infinitely differentiable), I may differentiate analytic function until it is zero."

    It is true that an analytic function is infinitely differentiable but what do you mean by "until it is zero"?

    You may be thinking of polynomials. All polynomials are infinitely differentiable and every time you differentiate you reduce the degree by one so if you differentiate a polynomial of degree n n-times it reduces to a constant and then the n+1 derivative (and all succeeding derivatives) are identically 0.

    But that is not true for general analytic functions. The function f(z)= e^z is analytic and so infinitely differentiable and all derivatives are e^z. Also y= cos(z) and sin(z) are analytic and all derivatives are one of cos(z), sin(z), -cos(z), -sin(z).
    So sorry, that what I've thought. I think about polynomials and forget about exponential. Then, can I conclude that the first statement is false. If it is false, which example should be an evidence that it is false??
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  12. #12
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    noppawit:

    Statement 1 is true, but your commentary underneath is not true in general; in other words, I think you may be misinterpreting the statement. If you are being asked to prove Statement 1, then I have sketched a proof in my above post. If you believe for some other reason that you can use Statement 1 to show eventually-vanishing derivitives for a general analytic function, then you are misguided.

    Let us know exactly what you are being asked to prove with regard to Statement 1.
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