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Degenerate Laurent Series

  1. May 4, 2012 #1
    Suppose we have

    [tex]f(z) = \sum_{n=-\infty}^\infty c_n z^n; \quad U(z) = \sum_{n=0}^\infty c_n z^n; \quad L(z) = \sum_{n=1}^\infty c_{-n} z^{-n}[/tex]
    where [itex]f(z)[/itex] converges* in the interior of some annulus with inner radius [itex]r[/itex] and outer radius [itex]R > r[/itex]. Further suppose [itex]U(z)[/itex] has radius of convergence [itex]R_0[/itex] and [itex]L(1/z)[/itex] has radius of convergence [itex]1/r_0[/itex], so [itex]L(z)[/itex] converges for [itex]|z| > r_0[/itex]. Must it be the case that [itex]r_0 \leq r < R \leq R_0[/itex]? Obviously the sum over all integers converges for [itex]r_0 < |z| < R_0[/itex] (if any such z exist), but for instance what happens if [itex]R_0 < r_0[/itex]? It's conceivable that the upper and lower sums both individually diverge, but that their combined sum still converges. It doesn't seem possible to me for this case to actually occur on a non-empty open set like an annulus... but I dunno.

    For instance, where does
    [tex]... + \frac{1}{2^3} \frac{1}{z^3} + \frac{1}{2^2} \frac{1}{z^2} + \frac{1}{2} \frac{1}{z} + 1 + z + z^2 + z^3 + ...[/tex]

    converge?

    This question came about in the context of the uniqueness of Laurent series. It seems that Laurent series are typically assumed to only be defined where their upper and lower sums converge, which sidesteps this issue. Each source I've looked at (Rudin's Real and Complex Analysis; Wikipedia; MathWorld; this book) at best includes this requirement implicitly and at worst ignores it entirely which makes me wonder if it's really necessary.


    *I'm interpreting [itex]\sum_{n=-\infty}^\infty c_n z^n[/itex] as [itex]\lim_{N \rightarrow \infty} \sum_{n=-N}^N c_n z^n[/itex].
     
  2. jcsd
  3. May 4, 2012 #2
    Yes.

    You can prove it playing a bit with power series. For example, take any [itex]z_0[/itex], [itex]r<|z_0|<R[/itex]. This implies that [itex]c_n|z_0|^n \to 0[/itex] as [itex]n\to \pm\infty[/itex], so the sequence [itex]c_n|z_0|^n [/itex] is bounded.

    Comparison with geometric series gives that [itex]L(z)[/itex] converges for all [itex]|z|> |z_0|[/itex]. Since [itex]U(z) =f(z) - L(z)[/itex], then [itex]U(z) [/itex] converges for all [itex]z[/itex], [itex]|z_0|< |z|<R[/itex] as difference of 2 convergence series (both f(z) and L(z) converge in this annulus).

    But power series (with positive powers) converge in a disc, so U(z) converges for [itex]|z|<R [/itex]. Then [itex]L(z) =f(z) - U(z)[/itex] converges in the annulus [itex]r< |z|<R[/itex] (f(z) converges in the annulus, L(z) converges in the disc [itex]|z|<R [/itex]). Then using comparison test we can concluyde that [itex]L(z) [/itex] converges for all [itex]|z|>r [/itex].
     
  4. May 4, 2012 #3
    I don't see how this is true, though you made me consider the fact that the nth term goes to 0 on this annulus, which gives the result anyway.

    Outside the radius of convergence of a power series, the nth terms do not go to 0 (which is how divergence is proven there), so each [itex]r < z < R[/itex] is not outside the radius of convergence of [itex]U(z)[/itex], so [itex]R_0 \ge R[/itex]. The lower bound is similar.
     
  5. May 5, 2012 #4
    It occurred to me there are some details missing yet. Hawkeye's statement, essentially
    on the annulus of convergence, requires some justification. The nth term test only gives
    [tex]\lim_{n \rightarrow \infty} c_n z^n + c_{-n} z^{-n} = 0 \mbox{ for } r < |z| < R \quad \mbox{ (*)}[/tex]
    which is not a priori the same. Suppose for now that [itex]|z| = 1[/itex] satisfies the above. Putting [itex]z=1, i[/itex] into (*) and simplifying gives
    [tex]\lim_{k \rightarrow \infty} c_{2k+1} + c_{-2k-1} = \lim_{k \rightarrow \infty} c_{2k+1} - c_{-2k-1} = 0[/tex]
    so that [itex]\lim_{k \rightarrow \infty} c_{2k+1} = 0[/itex]. Thus [itex]\liminf_{n \rightarrow \infty} |c_n| = 0[/itex]. Suppose to the contrary [itex]\limsup_{n \rightarrow \infty} |c_n| > 0[/itex]. There is some subsequence [itex]\{|c_{n_k}|\}[/itex] where [itex]\lim_{k \rightarrow \infty} |c_{n_k}| = a > 0[/itex]. Cutting off finitely many terms, this constrains [itex]\{c_{n_k}\}[/itex] to lie on a compact annulus not containing the origin, so the subsequence has a limit point [itex]c[/itex] in this annulus. Passing to a further subsequence, we may then take
    [tex]\lim_{k \rightarrow \infty} c_{n_k} = c \ne 0[/tex]
    Applying (*) with this subsequence and [itex]z=1[/itex] gives [itex]\lim_{k \rightarrow \infty} c_{n_k} + c_{-n_k} = 0 \Rightarrow \lim_{k \rightarrow \infty} c_{-n_k} = -c[/itex]. Taking [itex]z = e^{it}[/itex] in (*) gives
    [tex]\lim_{k \rightarrow \infty} c_{n_k} e^{i n_k t} + c_{-n_k} e^{-i n_k t} = 0[/tex]
    [tex]\Rightarrow \lim_{k \rightarrow \infty} c_{n_k} e^{2i n_k t} + c_{-n_k} = 0[/tex]
    [tex]\Rightarrow \lim_{k \rightarrow \infty} c_{n_k} e^{2i n_k t} = -(-c) = c[/tex]
    [tex]\Rightarrow \lim_{k \rightarrow \infty} e^{2i n_k t} = c/c = 1[/tex]
    This last line is obviously nonsense, for instance with [itex]t=\pi/2[/itex], so [itex]\lim_{n \rightarrow \infty} c_n = 0 = \lim_{n \rightarrow \infty} c_n z^n[/itex] after all, which forces [itex]\lim_{n \rightarrow \infty} c_{-n} z^{-n} = 0[/itex]. We may peel back the [itex]|z| = 1[/itex] assumption by applying the above to [itex]z / |z|[/itex].
     
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