Series Convergence

Fall 2020.2 #complex/qual/completed

problem (?):

Expand 11z2+1z3 in a series of the form anzn so it converges for

  • |z|<1,

  • 1<|z|<3,

  • |z|>3.

solution:

General strategy: each has two expansions, so just compute them all and pick appropriate ones for regions afterwards.

For 1z3: 1z3=1311z3=13k03kzk|z|<3=1z113z=z1k03kzk|z|>3.

For 11z2: 11z2=k0z2k|z|<1=1z211z2=z2k0z2k|z|>1.

So take 0<|z|<1f(z)=k0z2k13k03kzk1<|z|<3f(z)=z2k0z2k13k03kzk3<|z|<f(z)=z2k0z2k+z1k03kzk.

Spring 2020 HW 2.2 #complex/qual/completed

problem (?):

Let f be a power series centered at the origin. Prove that f has a power series expansion about any point in its disc of convergence.

concept:

    
  • Cauchy’s integral formula: f(z)=f(ξ)ξzdξ.
solution:

Idea: use Cauchy’s integral formula to get a series in (zz0). f(z)=f(ξ)ξzdξ=f(ξ)(1ξ(zz0)z0)dξ=f(ξ)ξz0(11w)dξw:=zz0ξz0=f(ξ)ξz0k0wkdξ=k0(f(ξ)ξz0dξ)wk=k0(f(ξ)ξz0dξ)wk=k0(f(ξ)(ξz0)k+1dξ)(zz0)k, where we’ve integrated over a curve contained in D the disc of convergence, and that the power series for f converges uniformly on D to commute the sum and integral.

Fall 2015, Spring 2020 HW 2, Ratio Test #complex/qual/work

problem (?):

Let an0 and show that lim

In particular, this shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series.

Analytic on circles #complex/qual/completed

problem (?):

Suppose f is analytic on a region \Omega such that {\mathbb{D}}\subseteq \Omega \subseteq {\mathbf{C}} and f(z) = \sum_{n=0}^\infty a_n z^n is a power series with radius of convergence exactly 1.

  • Give an example of such an f that converges at every point of S^1.

  • Give an example of such an f which is analytic at 1 but \sum_{n=0}^\infty a_n diverges.

  • Prove that f can not be analytic at every point of S^1.
solution:

    

Part a: Take f(z) \coloneqq\displaystyle\sum n^{-2}z^n, which converges absolutely for {\left\lvert {z} \right\rvert}=1 by the comparison test.

Part b: Take f(z) \coloneqq{1\over 1+z} = \sum_{k\geq 0} (-1)^k z^k, then f(1) = 2 by analytic continuation of the series at z=1. Then a_k = (-1)^k,

Part c: ??? Not clear if this is true, take f(z) = \sum z^n/n^2.

Spring 2020 HW 2.3: series on the circle #complex/qual/completed

problem (?):

Prove the following:

  • \sum_{n} nz^n does not converge at any point of S^1

  • \sum_n {z^n \over n^2} converges at every point of S^1.

  • \sum_n {z^n \over n} converges at every point of S^1 except z=1.
concept:

    
  • Summation by parts: Set B_0 \coloneqq 0, B_n \coloneqq\sum_{k\leq n} b_k, then \begin{align*} \sum_{n=M}^{N} a_{n} b_{n}=a_{N} B_{N}-a_{M} B_{M-1}-\sum_{n=M}^{N-1}\left(a_{n+1}-a_{n}\right) B_{n} .\end{align*}

  • Summing a geometric series: \begin{align*} \sum_{1\leq k \leq N} z^k = {1 - z^{N+1}\over 1-z} .\end{align*}

solution:

Part 1: This series does not have small tails: writing c_n \coloneqq n z^n we have {\left\lvert {c_n} \right\rvert} = {\left\lvert {nz^n} \right\rvert} = {\left\lvert {n} \right\rvert}\to \infty when {\left\lvert {z} \right\rvert} = 1.

Part 2: This converges absolutely and absolute convergence implies convergence: \begin{align*} {\left\lvert {\sum n^{-2} z^n} \right\rvert} \leq \sum {\left\lvert {n^{-2}z^n} \right\rvert} = \sum n^{-2} < \infty .\end{align*}

Part 3: Write f(z) = \sum_{k\geq 1} k^{-1}z^k. The value f(1) is the harmonic series, which we know diverges from undergraduate Calculus. For z\neq 1, apply summation by parts with a_k \coloneqq k^{-1} and b_k \coloneqq z^k, so

  • a_N = N^{-1}
  • a_M = M^{-1}
  • B_N = \sum_{k\leq N} z^k = {1-z^{N+1} \over 1-z}
  • B_M = \sum_{k\leq M} z^k
  • a_{n+1} - a_n = (n+1)^{-1}+ n^{-1}= - (n(n+1))^{-1}

Note that {\left\lvert {B_N} \right\rvert} \leq C_z \coloneqq{2\over {\left\lvert {1-z} \right\rvert} } for any N, since {\left\lvert {z} \right\rvert} = 1 is on S^1 and the maximum distance between two points on S^1 is 2. Moreover C_z < \infty when z\neq 1.

Applying the formula:

\begin{align*} {\left\lvert {\sum_{n=M}^N n^{-1}z^n } \right\rvert} &\leq {\left\lvert { N^{-1}B_N - M^{-1}B_{M-1} - \sum_{n=M}^{N-1} \left[ -(n(n+1))^{-1}B_n \right] } \right\rvert}\\ &\leq N^{-1}C_z + M^{-1}C_z + \sum_{M\leq n \leq N-1} C_z \qty{1\over n^2 + n}\\ &\leq C_z\qty{N^{-1}+ M^{-1}+ \sum_{M\leq n \leq N-1} n^{-2}} \\ &\overset{M, N\to\infty}\longrightarrow 0 ,\end{align*}

where we’ve used the triangle inequality and convergence of \sum n^{-2}. By the Cauchy criterion for sums, f(z) converges pointwise for z\neq 1.

Uniform convergence of series #complex/qual/work

problem (?):

Suppose \sum_{n=0}^\infty a_n z^n converges for some z_0 \neq 0.

  • Prove that the series converges absolutely for each z with {\left\lvert {z} \right\rvert} < {\left\lvert {z} \right\rvert}_0.

  • Suppose 0 < r < {\left\lvert {z_0} \right\rvert} and show that the series converges uniformly on {\left\lvert {z} \right\rvert} \leq r.

Sine series? #complex/qual/work

problem (?):

Prove that the following series converges uniformly on the set \left\{{z {~\mathrel{\Big\vert}~}\Im(z) < \ln 2}\right\}: \begin{align*} \sum_{n=1}^\infty {\sin(nz) \over 2^n} .\end{align*} Suppose 0 < r < {\left\lvert {z_0} \right\rvert} and show that the series converges uniformly on {\left\lvert {z} \right\rvert} \leq r.

Fall 2015 Extras #complex/qual/work

Assume f(z) is analytic in {\mathbb D} and f(0)=0 and is not a rotation (i.e. f(z) \neq e^{i \theta} z). Show that \displaystyle \sum_{n=1}^\infty f^{n}(z) converges uniformly to an analytic function on compact subsets of {\mathbb D}, where f^{n+1}(z) = f(f^{n}(z)).

#complex/qual/completed #complex/qual/work