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Math Help - Proof that an infinite series is analytic

  1. #1
    MHF Contributor arbolis's Avatar
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    Proof that an infinite series is analytic

    Prove that the function f(z)=1+ \sum _{n=1}^{\infty} \frac{a(a-1)...(a-n+1)}{n!}z^n is analytic when |z|<1 and that its derivative is \frac{a f(z)}{1+z}. Then demonstrate that the derivative of (1+z)^{-a} f(z) is worth 0 and deduce that f(z)=(1+z)^a. Note that (1+z^a)=exp(a Log(1+z)).

    My attempt: f(z)= \sum _{n=1}^{\infty} \frac{(a-0)(a-1)(a-2)...[a-(n-1)]}{n!} z^n=az+\frac{a(a-1 z^2)}{2}+\frac{a(a-1)(a-2) z^3}{3}+...=\alpha z + \beta z^2 + \gamma z ^3 +... where \alpha, \beta, \gamma,... \in \mathbb{R}. So it seems to me that f is analytic for any value of z though I don't know how to formally prove it. So I don't understand why they bother with |z|<1.
    I'd like to tackle this exercise and eventually finish it with your help. Thanks for any comment/tip.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Prove that the function f(z)=1+ \sum _{n=1}^{\infty} \frac{a(a-1)...(a-n+1)}{n!}z^n is analytic when |z|<1 and that its derivative is \frac{a f(z)}{1+z}. Then demonstrate that the derivative of (1+z)^{-a} f(z) is worth 0 and deduce that f(z)=(1+z)^a. Note that (1+z^a)=exp(a Log(1+z)).

    My attempt: f(z)= \sum _{n=1}^{\infty} \frac{(a-0)(a-1)(a-2)...[a-(n-1)]}{n!} z^n=az+\frac{a(a-1 z^2)}{2}+\frac{a(a-1)(a-2) z^3}{3}+...=\alpha z + \beta z^2 + \gamma z ^3 +... where \alpha, \beta, \gamma,... \in \mathbb{R}. So it seems to me that f is analytic for any value of z though I don't know how to formally prove it. So I don't understand why they bother with |z|<1.
    I'd like to tackle this exercise and eventually finish it with your help. Thanks for any comment/tip.

    Put a_n:=\frac{\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-n+1)\,z^n}{n!} , then after making some algebra we get \left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\alph  a -n}{n+1}\right||z|\xrightarrow [n\to\infty]{}|z| , and

    then the series converges absolutely for |z|<1 and it's thus an analytic function there.

    Tonio
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  3. #3
    MHF Contributor arbolis's Avatar
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    Oh thanks a lot. I will try to do the algebra on my own. So you applied the root test and saw that the radius of convergence depends on |z|. And if |z|<1 then the series converges absolutely and the function is therefore analytic.
    Is there a theorem that states that any absolutely convergent power series is an analytic function? I personally don't find it "that obvious".
    Correct me if I'm wrong: Any analytic function can be written as an infinite series. Must these infinite series converge absolutely? Or maybe a simple convergence is enough for some series to be analytic functions?
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