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Power series (part 2)

  1. Jul 21, 2014 #1
    Hi.
    I have another question about power series. I am having problem with the summarizing of the sum (writing in $\sum_{}^{}$ form).

    Here is the question:
    Let α be a real number that is not 0.
    Let $$f(z)=e^{{\alpha}Ln(z+1)}$$
    For integer n>0, find $$f^n(0).$$


    My partial solution:
    $$f(z)=e^{{\alpha}Ln(z+1)}=(z+1)^{\alpha}$$

    $$f^1(z)=α(z+1)^{α-1}$$

    $$f^2(z)=α(\alpha-1)(z+1)^{α-2}$$

    $$f^3(z)=α(α-1)(α-2)(z+1)^{α-3}$$

    The answer for $$f^n(z)$$ is

    $$f^n(z)=\alpha(\alpha-1)...(\alpha-n+1)(z+1)^{\alpha-n}.$$

    I am confused because when n=1,
    $$f^1(z)=\alpha(z+1)^{\alpha-1}$$
    but in the answer, there is the coefficient (α-1), this is where I am confused.
     
    Last edited: Jul 21, 2014
  2. jcsd
  3. Jul 21, 2014 #2

    CAF123

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    Did you make a typo in writing down the answer? There does not seem to be any ##n## index on the right hand side.
     
  4. Jul 21, 2014 #3
    You're right, there is a typo error. Give me a few minutes to correct it.
     
  5. Jul 21, 2014 #4
    $$f^n(z)=\alpha(\alpha-1)...(\alpha-n+1)(z+1)^{\alpha-n}$$
     
  6. Jul 21, 2014 #5

    CAF123

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    Yes, that's better. I think maybe it is just the notation that is confusing you. The quantity ##\alpha (\alpha - 1) \dots (\alpha - (n-1))## just means you write out all terms starting from ##\alpha## and ending with the term corresponding to that particular of ##n##. Putting a factor of ##\alpha - 1## in the general formula just serves to tell you that by increasing ##n## from 1 to 2, you should put in another factor of ##\alpha## lowered by 1. So, by extrapolating this knowledge to the general case, you see that ##f^n(z)## corresponds to the case where you have decreased ##\alpha## by 1 ##n-1## times and multiplied together all the consecutive terms.
     
  7. Jul 21, 2014 #6
    Ah, now I get why it's written that way, it's because it is corresponding to n. Thank you for explaining that. But I don't get this part, though. What do you mean by this:

    So, by extrapolating this knowledge to the general case, you see that ##f^n(z)## corresponds to the case where you have decreased ##\alpha## by 1 ##n-1## times and multiplied together all the consecutive terms.
     
  8. Jul 21, 2014 #7

    CAF123

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    By computing the ##n^{\text{th}}## derivative of ##f(z)##, you have decreased the value of ##\alpha## by ##n-1## times, right? E.g for n=2, you have factors ##\alpha## and ##(\alpha - 1)##. When I say 'multiply together all consecutive terms', I mean the general form for the ##n^{\text{th}}## derivative includes an overall term with all those factors multiplied together, e.g for n=2, you have a term ##\alpha(\alpha-1)##.
     
  9. Jul 21, 2014 #8
    Got it. So, the final answer is just plug in 0 into the function, right?

    ##∴f^n(0)= α(α-1)...(α-n+1)##
     
  10. Jul 21, 2014 #9

    CAF123

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    Right.
     
  11. Jul 21, 2014 #10
    Thanks. Problem solved.
     
  12. Jul 21, 2014 #11
    Hey, just one last question.

    In the question, ##f(z)=e^{{\alpha}Ln(z+1)}## and I have written it as ##f(z)=(z+1)^{\alpha}##.
    This is correct, right? Because someone said to me that it's not. He said:
    If z is not a real number, ##e^{\alpha \, \mathrm{Ln}\,\left( z + 1 \right) } \neq \left( z + 1 \right) ^{\alpha }## as the logarithm laws in general do not carry from the real numbers to the complex numbers...
     
  13. Jul 21, 2014 #12

    Ray Vickson

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    It is OK for the "principle branch" of the function ##\ln##. What you did is valid as long as ##z+1## stays away from the negative real axis; certainly it is OK if ##\text{Re}(z+1) > 0##, or ##\text{Re}(z) > -1##.
     
  14. Jul 21, 2014 #13

    HallsofIvy

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    You understand that [itex]e^{\alpha ln(z+ 1)}= (z+ 1)^\alpha[/itex], right?
     
  15. Jul 21, 2014 #14
    Yes i do.
     
  16. Jul 22, 2014 #15
    There's actually a continuation to this question. I have to expland f(z) into maclaurin series and then find the radius of convergence. To expand it, we have to use the exponential function method, right? Which is:
    $$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+\frac{z^2}{2!}+\frac{z^3}{3!}+...$$

    In this problem ##f(z) = e^{αLn(z+1)}##, how can I expand it into maclaurin series? And, how do I find the radius of convergence?
     
    Last edited: Jul 22, 2014
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