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Complex Analysis: Series Convergence

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    For ##|z-a|<r## let ##f(z)=\sum_{n=0}^{\infty}a_n (z-a)^n##. Let ##g(z)=\sum_{n=0}^{\infty}b_n(z-a)^n##. Assume ##g(z)## is nonzero for ##|z-a|<r##. Then ##b_0## is not zero.
    Define ##c_0=a_0/b_0## and, inductively for ##n>0##, define
    $$
    c_n=(a_n - \sum_{j=0}^{n-1} c_j b_{n-j})/b_0
    $$

    Note that the definition of ##c_n## implies that ##a_n=\sum_{j=0}^{n} c_j b_{n-j}## (it is equivalent to say ##c_n## solves this last equality)

    So, we have a formal series (no claim yet on converging to ##f/g##), ##\sum_{n=0}^{\infty} c_n (z-a)^n##. Call this the formal quotient.

    Take the formal quotient for granted. Prove that the formal quotient actually converges to ##f/g## for all ##|z-a|<r##


    2. Relevant equations


    3. The attempt at a solution
    I'm stumped. Any ideas on how to get started showing this?
     
  2. jcsd
  3. Mar 5, 2015 #2

    wabbit

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    What does "the formal quotient converges" (for a given value of z) mean ? (this is not a trick question)
     
  4. Mar 5, 2015 #3
    The limit of the partial sums converge?
     
  5. Mar 5, 2015 #4

    wabbit

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    Right. So did you have a look at the partial sums?
    Hint: don't think quotient, think product.
     
  6. Mar 5, 2015 #5
    Let ##q_N## be the partial sum of the formal quotient given in the problem and let ##f_N##, ##g_N## and ##r_N## be the partial sums of the taylor series for f,g and the remainder of their quotient respectively. Then I want to show
    $$\lim_{N\to\infty}(f_N=q_Ng_N+r_N)<\infty$$?

    Say I can show that everything to to the right of the equal sign inside the limit converges. That just shows that ##f_N## converges which I already know. I'm missing something.
     
  7. Mar 5, 2015 #6

    wabbit

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    What you need to do is look at the actual expressions for the partial sums. Write down the product of the partial sums for g an q and compare to the partial sum for f. Use the same number of terms for q and g but feel free to vary the number for f.

    Once you've done that it will look like almost a solution but not quite, and you're probably going to hit a roadblock, so I'll give you another hint to use then:
    Think about the radius of convergence.
     
  8. Mar 5, 2015 #7
    I get
    ##f_N=q_Ng_N=##
    ##\implies \sum_{n=0}^N a_n(z-a)^n=\sum_{n=0}^N c_n(z-a)^n \sum_{n=0}^N b_n(z-a)^n##
    ##\implies \sum_{n=0}^N a_n(z-a)^n=\sum_{n=0}^N (a_n/b_0-\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0})(z-a)^n \sum_{n=0}^N b_n(z-a)^n##
    ##\implies \sum_{n=0}^N a_n(z-a)^n=\sum_{n=0}^N \frac{a_n}{b_0}(z-a)^n-\sum_{n=0}^N \bigg(\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0}\bigg)(z-a)^n \sum_{n=0}^N b_n(z-a)^n##
    ##\implies \sum_{n=0}^N a_n(z-a)^n=f_N/b_0-\sum_{n=0}^N \bigg(\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0}\bigg)(z-a)^n \sum_{n=0}^N b_n(z-a)^n##
    and this ##\sum_{n=0}^N \bigg(\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0}\bigg)(z-a)^n## almost looks like ##q_Ng_N##
     
  9. Mar 5, 2015 #8

    wabbit

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    OK you are getting closer, this is the sort of manipulation you need to do to start.
    A few minor details just to make things easier. :
    - let's take a=0, it just makes the formulas simpler with nothing really lost and you can easily generalize at the end.
    - just work with the q.g product, leave aside the "f=" part
    - Don't use the formula expressing c_n, it's easier to just use the one for a_n=...
    These are not so important, its just for clarity really.

    Now one that is important : be careful with the ranges of your indices. The equation you arrive at isn't actually correct because of that.

    So. Let's start again. We'll call the partial sums F(N), G(N) and Q(N) with the obvious notations.

    You want to compute the product G(N)Q(N). Expand that and compare it with both F(N) and F(2N). (In your post you are trying to prove F(N)=G(N)Q(N), but that is not actually true - close but not quite, as you saw.)

    It will help if you draw a grid for the indices in each sum and carefully look at which terms (which indices) are involved in each expression.

    Once you do that you get to the roadblock I mentionned before.

    At that point maybe a short detour will be best : for a brief moment, we shall assume that all the coefficients as well as the variable z are positive real numbers. Do you see how to conclude in this special case ?

    Edit : I'm going to sign off pretty soon for today, so work on this and I'll check tomorrow to see if you still need more help.
     
    Last edited: Mar 5, 2015
  10. Mar 5, 2015 #9
    Note: I'm just saving my progress so far. I still need to work on the stuff below this note some more. Hope that's OK.

    ##q_Ng_N=\sum_{n=0}^N c_n(z)^n \sum_{n=0}^N b_n(z)^n##
    ##=\sum_{n=0}^N (a_n/b_0-\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0})(z)^n \sum_{n=0}^N b_n(z)^n##
    ##=\sum_{n=0}^N \frac{a_n}{b_0}(z)^n-\sum_{n=0}^N \bigg(\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0}\bigg)(z)^n \sum_{n=0}^N b_n(z)^n##
    ##=\bigg(\sum_{n=0}^N \frac{a_n}{b_0}(z)^n-\sum_{n=0}^N \frac{a_{n-1}}{b_0}(z)^n\bigg) \sum_{n=0}^N b_n(z)^n##

    Hmm, upon examining the last line, after I clean it up it looks like ##f_N/g_N## is some sort of recursive function of ##f_N##, which would converge since $$f$$ converges.
     
    Last edited: Mar 5, 2015
  11. Mar 6, 2015 #10
    ##q_Ng_N=\sum_{n=0}^N c_n(z)^n \sum_{n=0}^N b_n(z)^n##
    ##=\bigg(\sum_{n=0}^N (a_n/b_0-\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0})(z)^n\bigg) \sum_{n=0}^N b_n(z)^n##
    ##=\bigg[\sum_{n=0}^N \frac{a_n}{b_0}(z)^n-\sum_{n=0}^N \bigg(\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0}\bigg)(z)^n \bigg]\sum_{n=0}^N b_n(z)^n##
    ##=\bigg(\sum_{n=0}^N \frac{a_n}{b_0}(z)^n-\sum_{n=0}^N \frac{a_{n-1}}{b_0}(z)^n\bigg) \sum_{n=0}^N b_n(z)^n##
    ##=\frac{1}{b_0}\sum_{n=0}^N\bigg( a_n(z)^n-a_{n-1}(z)^{n-1}\bigg) \sum_{n=0}^N b_n(z)^n##
    ##=\frac{1}{b_0}[-a_0+a_{N-2}+a_{N}]g_N##
    ##\implies q_N=\frac{1}{b_0}[-a_0-a_{N-2}+a_{N}]z^n##
    If you let ##n## run to ##\infty## then you get
    ##\implies q_N=\frac{-a_0}{b_0}z^{-1}## since everything inside the region of convergence would cancel except the first term

    Edit: Darn, I'm still not quite there. Lot's of mistakes on the indices. I'll have to work on it more tomorrow.
     
    Last edited: Mar 6, 2015
  12. Mar 6, 2015 #11

    wabbit

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    You're making it more difficult than it needs to be. Instead of
    ##q_Ng_N=\sum_{n=0}^N c_n(z)^n \sum_{n=0}^N b_n(z)^n=\bigg(\sum_{n=0}^N (a_n/b_0-\sum_{j=0}^{N-1}\frac{c_jb_{n-j}}{b_0})(z)^n\bigg) \sum_{n=0}^N b_n(z)^n=...##
    Just use
    ##q_Ng_N=\sum c_n z^n \sum b_m z^m=\sum(\sum c_n b_m)z^p=...## together with ##\sum c_n b_m=...##

    As for the indices as I said use a drawing (grid of n=... x m=... ) and look at which terms go into ##f_N## and which go into ##f_{2N}##, it will be easier.

    Oh, and don't try going to th limit ##N\rightarrow\infty## too soon, you really need to get those partial sums straight first.
     
    Last edited: Mar 6, 2015
  13. Mar 8, 2015 #12
    This is probably obvious but just to be clear does ##f_{2N}=\sum_{n=0}^{2N}a_nz^n=a_0+a_1z^n+\cdots + a_{2N-1}z^{2N-1}+a_{2N}z^{2N}=f_N+\sum_{n=N+1}^{2N}a_nz^n## ?

    Also, I'm not 100% clear on what you mean by a "grid". However, I wrote out each term of ##\sum_{j=0}^{p}c_jb_{p-j}## thinking it would serve the same purpose. Does it?
     
  14. Mar 9, 2015 #13

    wabbit

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    Yes on both counts. The grid is just a visual aid : to me, the easiest way to compare which pairs of indices (j, p-j) to use notation in your last formula corespond to terms that go in fN, f2N, and qNgN , is to draw a grid. Otherwise I get a headache. But it achieves nothing you cannot do by careful examination of the formulas themselves. What works for you is what you should use.
     
  15. Mar 11, 2015 #14
    Here's an outline to a solution:
    ##f## and ##g## holomorphic implies ##\frac{f}{g}(z)=\sum_{n=0}^{\infty}c_n(z-a)^n## is holomorphic for ##|z-a|<r##

    By the product rule for power series
    ##f(z)=\frac{f}{g}(z)g(z)=\sum_{n=0}^{\infty}[\sum_{j=0}^{n}c_{j}b_{n-j}](z-a)^n##

    Then by uniqueness of the power series:
    ##a_n=\sum_{j=0}^n c_jb_{n-j}##
    now solve for ##c_n##...one method is to solve for ##c_0## then use induction.

    I left out some tedious to type details about disks of convergence
     
    Last edited: Mar 11, 2015
  16. Mar 13, 2015 #15

    wabbit

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    Sorry I missed your last post somehow. The thing is, those tedious details you left out about convergence are precisely the subject of the exercise. So no, this isn't an outline of a proof

    Alo it seems the hints I'm giving you don't help since you are not following them, and I'm unsure how to help in this circumstance - so perhaps you might want to try re-posting in a new thread so someone else might try another approach better suited to what you need.
     
    Last edited: Mar 13, 2015
  17. Mar 13, 2015 #16
    Sorry, wasn't trying to be rude. Your help was much appreciated and during the process I learned a lot. However, this solution is from the professor and was revealed before I was able to finish the problem using your strategy. For the sake of completeness I thought I should post it.

    Yes, the convergence portion of the problem is key but it wasn't the issue I was having. Perhaps that is why I carelessly called it a "tedious detail."

    Anyway, thanks again for the valuable insight you provided.
     
  18. Mar 13, 2015 #17

    wabbit

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    Oh OK sorry I misunderstood your post, it looked to me like you were suddenly trying a different direction and at the same time sweeping the key issue under the carpet:smile:
    (And you weren't rude at all, I was just at a loss as to how to proceed)
    Glad you sorted it out.
     
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